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13: Solutions and their Physical Properties

These are homework exercises to accompany the Textmap created for "General Chemistry: Principles and Modern Applications " by Petrucci et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.

Q13.1a

Which of the following do you except to be the least water soluble, and why? \(C_{10}H_{8(s)}\), \(NH_2OH_{(s)}\), \(C_6H_{6(l)}\),\(CaCO_{3(s)}\).

Q13.1a

\(CaCO_3\) least soluble because it does not interact with \(H_2O\) and the ion charge is significantly high.

Q13.1b

Which compound would be expected to readily dissolve in gasoline, and why?

CH3CH2OH(l), NH4 (aq), CH3(CH2)6COOH (s),BF3 (g)

Q13.1b

We would expect the caprylic acid (\(CH3(CH_2)_6COOH\))) to dissolve the easiest in gasoline because it is the only nonpolar molecule.

Q13.2a

Which of the following is not moderately soluble both in water and in benzene (\(C_6H_{6(l)}\)), and why? (a) 1-butanol, \(CH_3(CH_2)_2CH_2OH\); (b) naphthalene, \(C_{10}H_8\); (c) hexane,\(C_6H_{14}\) (d) \(NaCl_{(s)}\)

Q13.2a

  1. 1-butanol is soluble in water but not in benzene
  2. naphthalene is soluble in benzene but not in water
  3. hexane is soluble in benzene but not in water
  4. NaCl is soluble in water because it’s a ionic solid and the hydration energy is greater compared with the energy needed to separate ions from the ionic.

Q13.2b

What are some examples of heterogeneous and homogeneous mixtures?

S13.2b

Homogenous Mixture are the mixtures which have their components uniformly spread throughout the solution.

  • Examples: salt in water, sugar in water, 3 true solutions all are homogenous mixtures

Heterogeneous Mixture are the mixtures which have their components separately  throughout the solution.

  • Examples: sand in water, oil in water, all suspensions and colloidal solutions are heterogeneous mixtures

Q13.3

Substances that dissolve in water generally do not dissolve in benzene. Some substances are moderately soluble in both solvents, however. Which of those are substances? (a) para-Dichlorobenzene (b) Salicyl alcohol (c) Diphenyl (d) Hydroxyacetic acid

S13.3

  1. no
  2. yes because its OH and benzene ring
  3. no
  4. yes

Q13.31

A solution of 430.0g C7H16, 600.0g C5H12 and 150.0g C9H20 is prepared. What is the a) Mass percent, and b) mole percent of each component in the solution?

S13.31

a)

C7H16: (430.0gC7H16)/(600.0g+150.0g+430.0g) * 100%= 36.44%C7H16

C5H12: (600.0g C5H12)/(600.0g+150.0g+430.0g) * 100%= 50.85%C5H12

C9H20: (150.0g C9H20)/(600.0g+150.0g+430.0g) * 100%= 12.71%C9H20

 

b)       430.0g C7H16 * (1 moleC7H16)/(100.2g C7H16)= 4.29 mol C7H16

 

600.0g C5H12 * (1 moleC5H12)/(72.15gC5H12)= 8.32 mol C5H12

150.0g C9H20* (1 moleC9H20)/(128.26gC9H20)= 1.17 mol C9H20

(4.29 mol C7H16)/(4.29 mol C7H16+ 8.32 mol C5H12+1.17 mol C9H20)*100%=31.13%C7H16

(8.32 mol C5H12)/(4.29 mol C7H16+ 8.32 mol C5H12+1.17 mol C9H20)*100%=60.38%C5H12

(1.17 mol C9H20)/(4.29 mol C7H16+ 8.32 mol C5H12+1.17 mol C9H20)*100%=8.49%C9H20

Q13.33a

Calculate the mole fraction of solute for these substances.

  1. 1500 g H2O and 250 g NaCl            
  2. 230 g C2H5OH and 720 g H2O

S13.33a

a)23g/mol+35.5g/mol=58.5g/mol

 

250g/58.5=4.27mol NaCl

1500g/18g/mol=83.3mol H2O

NaCl= 4.27/4.27+83.3=.05

 

b)230g/46g/mol=5molC2H5OH

 

720g/18g/mol=40molH2O

mol fraction C2H5OH=5/40+4=1/9=.11

Q13.33b

Calculate the mole fraction of the solute in the following aqueous solutions: (a) 0.221M C6H1206 (d=3.20g/mL); (b) 5.1% ethanol, by volume (d=2.001g/mL); pure CH3CH2OH, d=0.989g/mL).

Q13.33b

(a)  Moles of C6H12O6= 1.00L*(0.221mol C6H12O6/1.00L)=0.221mol

C2H12O6

  • Mass of solution=1000mL Soln*(3.20g Soln/1.0mL Soln)=3200g Soln
  • Mass of C6H12O6=0.221mol C6H12O6*(180g C6H12O6/1mol C6H12O6)=39.78g
  • Mass of H20=3200-39.78=3160.22g \(H_2O\)
  • Moles of H20=3160.22g H20*(1mol H2O/18.02g H2O)=175.373mol \(H_2O\)

XC6H1206=0.221mol C6H12O6/(0.221mol+175.373mol)=.00126

(b)  Mass of ethanol=5.1ethanol*(0.989g ethonal/1.0mLethonal)=5.044g

Ethonal

  • Mass of Soln=100.0mL Soln*(2.001g Soln/1.0mL Soln)=200.1g
  • Mass of H20=200.1-5.044=195.056g \(H_2O\)
  • Mol C2H5OH=5.044*(1.0mol C2H5OH/46.07g C2H4OH)=0.109
  • Mol H20=195.056*(1/18.02)=10.824

XC2H5OH=0.109mol/(0.109mol+10.824mol)=0.001

Q13.35

What volume of glycerol, CH3CH(OH)CH2OH (d=3.02g/mL),must be added per kilogram of water to produce a solution with 5.50mol % glycerol?

Q13.35

Nwater=1000g H20*(1mol H2O/18.02g H2O)=55.49 mol \(H_2O\)

Xgly=0.0550=Ngly/Ngly+55.49 Ngly=0.550 Ngly+3.05

Ngly=3.05/(1.0000-0.0550)=2.88mol glycerol

Volume glycerol=2.88mol C3H8O3*(92.03g C3H8O3/1mol

C3H8O3)*(1mL/3.02g)=87.8mL glycerol

Q13.39a

Refer to Figure 13-8 and determine the molality of NH4Cl in a saturated aqueous solution at 50ºC.

Q13.39a

According to Figure 13-4, at 50ºC, solubility is expressed by about 51g of NH4Cl per 100g H20.

Therefore, through stoichiometry, molality = 9.6

(51gNH4Cl/100gH20)/(1mol/53.49gNH4Cl)(1000g/kg)=9.625m

Q13.39b

Refer to Figure 13-8 and determine the molality of \(NH_4Cl\) in a saturated aqueous solution at 60 C.

Solution:

At 60C the solubility of \(NH_4Cl\) is 56.3 g per 100 g of \(H_2O\).

\[Molality=(56.3\;g*\dfrac{(1\;mol\; NH_4Cl/53.49g NH_4Cl))}{(100\;g H_2O (1\;kg / 1000\;g)}=10.53\;m\]

Q13.41

A solution of 15.0g \(KClO_4\) in 450 g of water is brought to a temperature of 40C

  1. Refer to Figure 13-8 and determine whether the solution is unsaturated or supersaturated at 40C
  2. Approximately what mass of \(KClO_4\), in grams, must be added to saturate the solution (if originally unsaturated), or what mass of \(KClO_4\) can be crystallized (if originally supersaturated)?

S13.41

  1. mass solute/100g H2O=100 g H2O*(15.0g KClO4/450.0g water=3.33g KClO4. At 40c a saturated KClO4 solution has a concentration of about 4.6 g KClO4 dissolved in 100g water. Thus the solution is unsaturated.
  2. Mass of be added= (450g H2O*(4.6g KClO4/100g H2O)-15.0g KCIO4=20.7-15=5.7g KClO4

Q13.43a

If the Henry’s law constant of Nitrogen gas dissolved in water at 35 ºC is 6.40x10^-4 (mol/L) /atm in 3 L of water, how many grams of nitrogen gas is there under a pressure of 2 atm at 25 ºC?

Q13.43a

C=kp

C=(6.40 x 10-4 M/atm)(2atm)= .00128 M

(.00128 mol/L)(3L)= .00384 mol

.00384 mol (28.02 g/mol)= .11067 g N2.

Q13.43b

Under a pressure of 1.00 atm, 43.25 mL of \(O_{2(g)}\) dissolves in 2.3 L \(H_2O\) at 25 C. What will be the molarity of \(O_2\) in the saturated solution 25C when the \(O_2\) pressure is 5.49 atm?

Q13.43b

\[Molarity=(0.04325\;L \;O_2*(1\;mol\; O_2/\;24.465\;L\;O_2))/2.3\;L=7.69 \times 10^{-4}M\]

Henry’s law constant for \(O_2\):

\[\(K_H=C/P_{gas}=7.69 \times 10^{-4}M / 1.00 atm = 7.69 \times 10^{-4}\;M/atm\]

Henry’s Law C=KH*Pgas=(7.69*10^-4M/atm)*(5.49atm)=0.00421M

Q13.45

Natural gas consists of 90% methane, \(CH_4\). Assume that the solubility of natural gas at 20C and 1 atm gas pressure is about the same as that of \(CH_4\), 0.032 g/kg water. If a sample of natural gas under a pressure of 35 atm is kept in contact with 1.04*10^3kg of water, what mass of natural gas will dissolve?

S13.45

\[Mass\; of CH_4=1.04 \times 10^3\;kg *(0.032\;g/1\;kg H_2O\; atm) (35 \;atm)=1164\;g CH_4\]

Q13.47

The aqueous solubility at 20C of Ar at 1 atm is equivalent to 53.2 mL \(Ar_{(g)}\), measured at STP, per liter of water. What is the molarity of Ar in water that is saturated with air at 1 atm and 20C? Air contains 0.897% Ar by volume. Assume that the volume of water does not change when it becomes saturated with air.

Q13.47

\[K_{Ar}=C/P_{Ar}=((53.2mL Ar/1L soln)*(1 mol Ar/ 22,414 mL at STP))/2.4 atm pressure=0.0024\;M/atm\]

\[M=K_{Ar} \, P_{Ar}=(0.0024\;M/atm) (0.00897\;atm)=2.1 \times 10^{-5}M\]

Q13.49

Henry’s law can be stated this way: The mass of a gas dissolved by a given quantity of solvent at a fixed temperature is directly proportional to the pressure of the gas. Show how this statement is related to equation (13.2)

Q13.49

While the gas dissolves in liquid, the solution remains essentially constant because of the low density of molecules in the gaseous state. Changes in the number of dissolved gas molecules causes the changes in the concentrations in the solution. The number is directly proportional to the mass of dissolved gas.

Q13.51a

What is the vapor pressure (in mmHg) of a solution of 4.40 g of Br2 in 101.0 g of CCl4 at 300 K? The vapor pressure of pure bromine at 300 K is 30.5 kPa and the vapor pressure of CCl4 is 16.5 kPa.

S13.51a

1) Calculate moles, then mole fraction of each substance:

Br2 ⇒ 4.40 g / 159.808 g/mol = 0.027533 mol

CCl4 ⇒ 101.0 g / 153.823 g/mol = 0.6566 mol

χBr2 ⇒ 0.027533 mol / 0.684133 mol = 0.040245

χCCl4 ⇒ 0.6566 mol / 0.684133 mol = 0.959755

2) Calculate total pressure:

Ptotal = P°Br2χBr2 + P°CCl4χCCl4

x = (30.5 kPa) (0.040245) + (16.5 kPa) (0.959755)

x = 1.2275 + 15.8360 = 17.0635 kPa

3) Convert to mmHg:

17.0635 kPa x (760.0 mmHg / 101.325 kPa) = 128 mmHg

Q13.51b

What are the partial and total vapor pressures of a solution obtained by mixing 43.4 g benzene, \(C_6H_6\), and 65.3 g toluene, \(C_6H_5CH_3\), at 25C? The vapor pressure of \(C_6H_6\) at 25C is 95.1 mmHg; the vapor pressure of \(C_6H_5CH_3\) is 28.4 mmHg.

Q13.51b

\[N_1=(43.4\;\cancel{g}) \left(\dfrac{1\; mol}{78.11\; \cancel{g}}\right)=0.556\;mol\; C_7H_8\]

\[N_2=(65.3\;\cancel{g}) \left(\dfrac{1\; mol}{92.14\; \cancel{g}}\right)=0.709\;mol\; C_6H_6\]

\[X_1=\dfrac{0.556}{0.556+0.709}=0.440\]

\[X_2=\dfrac{0.709}{0.556 + 0.709}=0.560\]

\[P_1=(0.44)(95.1)=41.8\; mmHg\]

\[P_2=(0.56)(28.4)=15.9\; mmHg\]

\[P_{Total}=P_1+ P_2 = 41.8\;mmHg + 15.9\;mmHg =57.7\;mmHg\]

Q13.53

Calculate the vapor pressure at 25C of a solution containing 178g of the nonvolatile solute, glucose, \(C_6H_{12}O_6\), in 967g \(H_2O\). The vapor pressure of water at 25C is 23.8 mmHg.

Q13.53

Nglucose=178g*(1mol/180.2g)=0.988mol

Nwater= 967g*(1mol/18.02g)=53.7mol

Xglucose=nglucose/(nglucose+nwater)=0.988/(0.988+53.7)=0.0181

PA^0-PA/PA^0=Xsolute

(23.8mmHg-PA)/23.8mmHg=0.0181

PA=23.4mmHg

Q13.57

A benzene-toluene solution with Xbenz=0.308 has a normal boiling point of 98.6C. The vapor pressure of pure toluene at 98.6C is 533mmHg. What must be the vapor pressure of pure benzene at 98.6C?

S13.57

Mole fraction of Toluene Xt=1-Xb=1-0.308=0.692

Ptoluene=XtoluenePtoluene=0.692*533mmHg=369mmHg

Partial pressure of benzene in solution=760mmHg-369mmHg=391mmHg

Partial pressure of benzene=mole fraction*vapor pressure

391mmHg=0.308*Pb

Pb=391/0.308=1269.5mmHg

Q13.59a

A 0.63 g sample of polyvinyl chloride is dissolved is 324mL of a suitable solvent at 21C. The solution has an osmotic pressure of 1.67 mmHg. What is the molar mass of PVC?

S13.59a

N/V=Pi/RT=(1.67mmHg*(1atm/760mmHg))/0.08206Latmmol^-1K^-1*294.2K)=9.10*10^-5M

Solute Amount=0.324L*(9.10*10^-5mol/1L)=2.95*10^-5mol M=0.63g/2.95mol=21355.9g/mol

Q13.59b

What is the osmotic pressure of a 125mL solution containing 7.6g glucose at 37˚C?

S13.59b

\[π=iMRT\]

i=1because glucose does not dissociate into ions.

M= (7.6gC6H12O6)(1mol/180.16g)= 0.042mol/(0.125L)=0.336mol/L

R=0.08206L*atm*mol-1*K-1

T=37˚C+273.15=310.15K

=(0.336mol/L)(0.08206L*atm*mol-1*K-1)(310.15K)=8.55atm

Q13.61

The stem of cut flowers wilt when they are in NaCl(ap) concentration. Fresh cucumber shrivels are in similar situation. Explain the basis of these phenomena.

S13.61

Both of them have ionic solutions, but the solutions are less concentrated than the salt solution. The solution in the plant material moves across the semi permeable to dilute the salt solution, leaving behind flowers wilt and pickles shrivel.

Q13.63

In what volume of water must be 3mol of a nonelectrolyte be dissolved if the solution is to have an osmotic pressure of 4atm at 301K? Which of the gas laws does this result resemble?

Solution:

n/V=Pi/RT=(4atm/0.08206Latmmol^-1K^-1*301K)=0.162M

Volume=3mol*(1L/0.162mol solute/0=18.5L solvent

Osmotic Pressure Eq.

Q13.65

At 25C a 0.71g sample of polyisobutylene in 200.0mL of benzene solution has an osmotic pressure that supports a 9.1 mm column of solution (d=0.88g/mL.) What is the molar mass of the polyisobutylene? (Hg, d=13.6g/mL)

Solution:

Pi=9.1mm soln*(0.88mmHHg/13.6mm soln)*(1atm/760mmHg)=7.7*10^-4atm

n/V=pi/RT=7.7*10^-4atm/0.08206Latmmol^-1K^-1*298K=3.2*10^-5M

Amount solute=200.0mL*(1L/1000mL)*(3.2*10^-5M)=6.3*10^-6mol solute

Molar mass=0.71g/6.3*10^-6mol=110000g/mol

Q13.71

Determine the new freezing point of a solution made from 3kg of water and 2.5mol of CaCl2. Freezing point constant for water is -1.86ºC/m

S13.72

∆T=imKf    i=1

CaC2 → Ca2+ + 2Cl-

m=2.5mol/3kg=.833m

(3)(.833m)(-1.86ºC/m)=∆T=-4.65ºC

0-4.65ºC=-4.65ºC

Q13.75

Triophene (fp=-38.3;bp=84.4C) is a sulfur-containing hydrocarbon sometimes used as a solvent in place of benzene. Combustion of a 2.348g sample of thiophene produces 5.012g CO2, 1.150g \(H_2O\), and 1.788 g SO2. When a 0.867g thiophene is dissolved in 44.56g of benzene, the freezing point is lowered by 1.183C. What is the molecular formula of thiophene?

Q13.75

M=changeTf=-1.183C/-5.12C/m=0.131m

Amount of solute=0.04456 kg benzene*(0.231mol solute.1kg benzene)=0.0103mol solute

Molar mass=0.867g thiophene/0.0103mol thiophene=84.3g/mol

C=5.012g*(1mol CO2/44.010g)*(1mol C/1mol CO2)=0.1134mol/0.02791=4.08mol C

H=1.150g*(1mol H2O/18.015g H2O)*(2mol H/1mol H2O)=0.1277mol H/0.02791=4.575mol H

S=1.788g*(1mol SO2/64.065g SO4)*(1mol S/1mol SO4)=0.02791mol S/0.02791=1.000mol S

Q13.77

Cooks often add some salt to water before boiling it. Some people say this helps the cooking process by raising the boiling point of the water. Others say not enough salt is usually added to make any noticeable difference. Approximately how many grams of NaCl must be added to a liter of water at 1 atm pressure to raise the boiling point by 3.4C? Is this a typical amount of salt that you might add to cooking water?

S13.77

M=change/iK=3.4C/2*0.512C/m=3.3m

Solute mass=1.00l H20*(1kg H2O/1L H2O)*(3.3mol NaCl/1kg H2O)*(58.4g NaCl/1mol NaCl)=190g NaCl

This is at least ten times the amount of salt in a liter of water.

Q13.81

Predict the approximate freezing points of 0.10 m solutions of the following solutes dissolved in water:

  1. CO(NH2)2;
  2. NH4NO3;
  3. HCl;
  4. CaCl2;
  5. MgSO4;
  6. C2H5OH;
  7. HC2H3O2

S13.81

ΔT = Kf * m * i

 

a) ΔT = (1.86 °C/m)*(.2 m)*(1)=0.372

   freezing point = 0ºC-0.372ºC = 0.372ºC

b) ΔT = (1.86 °C/m)*(.2 m)*(2)=0.744

  freezing point = 0ºC-0.744ºC = 0.744ºC

c) ΔT = (1.86 °C/m)*(.2 m)*(2)=0.744

  freezing point = 0ºC-0.744ºC = 0.744ºC

d) ΔT = (1.86 °C/m)*(.2 m)*(3)=1.116

  freezing point = 0ºC-1.116ºC = 1.116ºC

e) ΔT = (1.86 °C/m)*(.2 m)*(2)=0.744

  freezing point = 0ºC-0.744ºC = 0.744ºC

f) ΔT = (1.86 °C/m)*(.2 m)*(1)=0.372

  freezing point = 0ºC-0.372ºC = 0.372ºC

g) ΔT = (1.86 °C/m)*(.2 m)*(1)=0.372

  freezing point = 0ºC-0.372ºC = 0.372ºC

Q13.83

NH3(aq) conducts weakly electric current. The same is true for acetic acid, HC2H3O2(aq). When the solutions go together, the resulting solution conducts very well electric current. Why?

S13.83

NH3 with HC2H3O2=NH4C2H3O2, solution of ions NH4+ and CH3COO-

NH3(aq)+HC2H3O(aq)àNH4C2H3O2(aq)

NH4C2H3O2(aq)àNH4+(aq)+C2H3O2-(aq)

This is a strong electrolytes that conducts strong currents

Q13.87

A typical root beer contains 0.11% of a 72% \(H_3PO_4\) solution by mass. How many milligrams of phosphorus are contained in a 13 oz (29.6 mL) can of this root beer? Solution density is \(\rho=1.00 \;g/mL\).

S13.87

Mass of root beer=13oz*(29.6mL/1oz)*(1.00g/1.00mL)=384.8g

Mass of 72%H3PO4 solution=(0.13/100)*384.8g= 0.500g

Mass of H3PO4=(72/100)*(0.500g)=0.360g

Mass % phosphorus in phosphoric acid=(mass of phosphorus/mass of phosphoric acid)*100=(30.974g/98.00g)*100=31.61

Mass of Phosphorus=(31.61/100)*0.360g=0.114g

Q13.88

An aqueous solution 113.1g KOH/L solution. The solution density is 1.11 g/mL. Your task is to use 100.0mL of this solution to 0.25m KOH. What mass of which component, KOH or \(H_2O\), would you add to the 100.0mL of solution?

Solution:

KOH molarity=(113.1g KOH*(1mol KOH/56.010g KOH))/1L soln=2.019M

\(H_2O\) in final soln=0.1000L orig.soln*(2.019mol KOH/1L soln)*(1kg H20/0.250mol KOH)=0.810kg \(H_2O\)

Mass of original solution 100.0mL*1.11g/mL=111g original solution

Mass KOH=100.0mL*(1L/1000mL)*(113.1g KOH/1Lsoln)=11.31g KOH

Original mass of water=113.1g soln-11.31g KOH=101.79g \(H_2O\)

Mass added H2O=810g H2O-101.79g H2O=678.21g \(H_2O\)

Q13.103

Suppose that 1.15mg of gold is obtained in a colloidal dispersion in which the gold particles are spherical, with a radius 1.00*10^2nm. Density=18.23g/cm^3) (a) What is the total surface area of the particles? (b) What is the surface area of a single cube of gold of mass 3.07mg?

S13.103

(a)Surface area=4pi^2=4(3.1416)(1*10^-7m)^2=1.26*10^-13m^3

Particle volume= 4pir^3/3=4pi(1*10^-7)^3/3=4.19*10^-21m^3

Particle mass=DV=(18.23g/cm^3)*(100cm)^3/(1m)^3*4.19*10^-21m^3=7.64*10^-14g/particle

Number of Au particle=mass of Au/particle mass=(1.00*10^-3g Au)/7.64*10^-14g/particle=1.309*10^10 particle

Total surface area=(1.26*10^-13m^2/particle)(1.309*10^10particle)=0.00162m^2

(b)Au: (3.07mg)(10^-3g/mg)/(18.23g/cm^3)=1.68*10^-4cm^3

      L=(1.68*10^-4cm^3)^(1/3)=0.0552cm

      Area=6*L^2=6*(0.0552cm)^2*(1m/100cm)^2=1.83*10^-8cm^2

Q13.113

What volume of ethylene glycol (\(HOCH_2CH_2OH\)) with density, \(\rho=2.21\; gm/L\), must be added to 21.12 L of water (\(K_f=1.91 \;C/m\)) to produce a solution that freezes at -15 C?

Solution:

\[\Delta T_f=-k_f m\]

\[T-T_f=-k_f m\]

-15C-0.00C=-1.91Cm^-1*m

molality=-15C/-1.91Cm^-1=7.85m

mol HOCH_2CH_2OH =(7.85mol/1kg H20)*(1kg H2O/1L H2O)*21.12L=165.86mol

Volume of ethylene glycol=165.86 mol C2H6O2*(62.09g C2H6H02/1mol C2H6H02)*(1mL/2.21g)*(1L/10^3mL)=4.66 L

Q13.117

What are the following terms or symbols: (a) Xb; (b) PA^0 (c) Kf (d) i

S13.117

  1. mole fraction of liquid B. Number of moles of B to the total number of moles of the solution
  2. vapor pressure of pure solvent at the given temp \[P_A^0=\dfrac{P_A}{X_A}\]
  3. molal depression constant or cryoscopic constant
  4. ratio of measured vaule of a colligate property to the excepted value of the solute in a non-electrolyte \[i=\dfrac{changeT_f}{excepted change T_f}\]

 

 

 

 

 

 

1)   _____________ ( CHCl2(l)/BF3/CCl4(l)(choose one) is expected to be the most water soluble because _________________.

 

2)   _____________ (C6H14/C6H5OH/CCl4/C10H8) (choose one) is moderately soluble in both water and benzene (C6H3Cl) because ________________.

 

3)   Most substances that are soluble in water aren’t soluble in benzene (and vice-versa). However, there are substances that are moderately soluble in both water and benzene, _________________,(CH4   / Ch3(CH2)2CH2OH) (choose one) is an example of this.

 

33) What is the mol fraction of the solute in the following?

         
a)a solution prepared by mxing 2.17 moles of C7H16, 1.5 moles C8H18, and 2,7 moles C9H20

 

35) Your friend wants to produce a solution that is 7.23% C3H8O2, what volume of, C3H8O2, would you suggest your friend to add per kilogram of water to achieve this?

 

39) if a 100ml sample of water at 293 K contains 13 ppm of aluminum, then how many aluminum ions are present in the solution and what is the molality of the solution?

 

41) Consider a solution consisting of 32g of KClO4 and 500 g of water that is heated to a temperature of 313.15K.

  a) Determine whether the solution is saturated, unsaturated, or supersaturated at 313.15 K.

 

43) 28.31 mL of CO2(g) dissolves in 1.0L water at 25˚C at a CO2 pressure of 1 atm, if the pressure rose to 4.2 atm (all other variables held constant) what will the molarity of CO2 in the saturated solution be?

 

45) What mass of natural gas will dissolve if a sample of natural gas is under a pressure of 17 atm is kept in contact with 1000 kg of water? Consider that the solubility of natural gas at 293 K and 1.0 atm is about 0.037g/Kg water.

 

47) Neon (Ne) at 1 atm has an aqueous solubility equal to 25.9 ml Ne(g) measured at STP/L water.  Determine molarity of Ne in water saturated with air at 1.0 atm and 293 K.  (air has .0015% Ne by volume.)

 

49) Using Henry’s law, describe why the pressure at a fixed temp can increase.

 

51) If one were to mix 40.3 g of C6H6 and 53.5 g of C6H5CH3 at 298 K then what will be the corresponding partial pressures? Total pressure?  (P˚C6H6 =95.1 torr, P˚C6H5CH3= 28.4 torr).

 

53) For a solution of .76 mol of NaCl in 690 g H2O what will be the vapor pressure at 298K? (P˚water= 23.8mmHg @ 25˚C)

 

55) The products of a reaction are 27% C6H5CH=CH2  and 73% C6H5CH2CH3 (by mass).  Given this, if the mixture were to be separated via fractional distillation at 363K, what would be the vapor pressures at equilibrium (P˚C6H5CH=CH2= 134mmHg, P˚C6H5CH2CH3= 182mmHg)?

 

57) Given a solution that is 40% benzene and 60% toluene with a boiling point of 371.6 K, what is P˚benz at 371.6K? (P˚toluene= 533mmHg @371.6K)

 

59) if a .58 g  sample of CO2 is dissolved in 250 ml of an appropriate solvent at 298K and the solution has an osmotic pressure of 2 mmHg then what is the molar mass of CO2?

 

61) Describe what occurs in terms of osmotic pressure when a cucumber is placed into a solution of highly concentrated salt and shrivels up.

 

63) what will be the osmotic pressure of a 2 M aqueous solution at 293 K?

 

65) If an aqueous solution has 0.97 g/L of an organic solution then the osmotic pressure of the solution will be 62.9 torr, at 298K. What is the molar mass (in g/mol) of this solution?

 

75) a compound is composed of approximately 40.3% B, 52.2% N, 7.5% H by mass.  When 2.8867g is dissolven in 50 g benzene the solution freezes at 1.3˚C, what is the molecular formula of the compound> (Fp pure benz=5.48˚c ;                                          Kb benz=5.12˚C/k)(density benz=.879 g/ml).

 

77)  what amount of NaCl must be added to a 2.37 sample of H2O at a pressure of 1 atm would be needed in order to ingrease the boiling point by 2˚C?

 

83)  predict the freezing points of the following .25 m solutions when dissolved in water (∆Tf=-ikfm            Kf water= 1.84˚C/m

 

a)           Co(NH2)2

b)           NH4NO3

c)    HCl

d)   CaCL2

 

 

87) Consider a solution (A) that has .617 g CO(NH2)2 dissolved in 90 g of water; and solution B has 3.7 g of C12H22O11 in 80 g of water.  What would be the compositions of each of these at equilibrium?

 

88) Given two isomers with differing freezing points, boiling points, and densities what methods could be instituted in order to separate the two if they are mixed in solution?

 

113) what will be the resulting vapor pressure when 58.9 g C6H14 is introduced into a container with 44 g of C6H6 at 332K if P˚ C6H14= 573mmHg and P˚ C6H6= 391 

S13.1

   CH2Cl2(l) is the most water soluble because water is nonpolar and of the molecules given CH2Cl2(l) is the least polar molecule.

2)   You need a polar molecule that is not ionic, phenol, C6H5OH, is the only choice that makes sense.

3)   Ch3(CH2)2CH2OH because the butyl chain is nonpolar while the OH groups are polar.

S13.33

2.17 moles of C7H16, 1.5 moles C8H18, and 2,7 moles C9H20

moltot=6.37

mol fracions:

C7H16=2.17/6.37= .34

C8H18=1.5/6.37= .235

C9H20= 2.7/6.37= .424

S13.35

      mols H2O= 1000g(1 mol/(18.02g/mol))=55.49 mols H2O

      X (C3H8O2)=7.23%=.0723=n(C3H8O2)/(n(C3H8O2)+nH2O)

      n(C3H8O2)=.073(n(C3H8O2))+4.05=4.37 mol (C3H8O2)

      4.37 mol (C3H8O2)*(92.09g/mol)*(1ml/1.26g)=

 

.319.39 ml  (C3H8O2) needed to produce the solution

S13.39

a)

 Al=101.996 g/mol

mol Al= (13ug/100g H2O)*(1g/106ug)*(1mol Al/101.96gAl)=1.275E-9 mols Al

# Al atoms= (1.275E-9)(6.022E23)=7.68E14 atoms

 

b)

molality:

            m=1.275E-9/(100gH2O*(1kg/1000g))=1.27E-8 m

 

41) [KClO4]= 100g  H2O * (32g/500g H2O)= 6.4 g KCLO4

                  At 313.15K, a saturated solution has a concentration of about 4.6g in 100g H2O.  Based on this the solution is supersaturated.

 

43) 28.31 mL of CO2(g) dissolves in 1.0L water at 25˚C at a CO2 pressure of 1 atm, if the pressure rose to 4.2 atm (all other variables held constant) what will the molarity of CO2 in the saturated solution be?

 

PV=nRT=> n=PV/RT=(1atm*.02831 L)/(.08206*298K)= .001158 mol CO2

[CO2]= .001158mol/1.0L soln= .001158 M

 

concentration at higher pressure:

[CO2]1=.001158*4.2atm= .00486 M

 

45)

mass of natural gas= 1000 kg H2O*(.037g/kg)*17 atm = 629 g natural gas dissolved.

 

47)

henry’s law: C=kP

 

k=C1/P1=C2/P2

kNe=C/PNe=((25.9ml/L)(1 mol Ne/22414 ml at STP))/1atm= .001155M/atm

 

C=KNePNe= (.001155M/atm)(.000015)=1.73E-8 M Ne

 

49) Due to the low density of gas molecules, the volume of a solution will remain the same as a gas is dissolved in the liquid. When this happens the concentration of gas increases, this is proportional to the mass of gas that is dissolved and will cause an increase in pressure.

 

51)

nC6H6= 40.3 g * (1 mol/78.11g)= .516 mols

nC6H5CH3= 53.5 g * (1 mol/ 92.14 g) = .580 mols

 

XC6H6= .516/(.516+.580)= .4708

XC6H5CH3=  .580/(.516+.580)= .5292

 

PC6H6= .4708*95.1 torr= 44.77 torr= 44.77 mmHg

PC6H5CH3= .5292*28.4 torr= 15.029 torr= 15.029 mmHg

 

Ptot= 44.7 + 15.029 =59.799 mmHg

 

53)

nNaCl=.76 mols (given)

nH2O= 690 g/(18.02 g/mol) = 38.29 mols

 

Xwater=(38.29/(38.29+.76))=.98

 

Psoln= XwaterPwater =(.98)(23.8mmHg)= 23.324 mmHg

 

55)

(27g(C6H5CH=CH2)/(104g/mol))= .26 mols

(73g(C6H5CH2CH3)/(106g/mol))= .69 mols

 

X(C6H5CH=CH2)= .26/(.26+.69) = .27

P(C6H5CH=CH2)= (.27)(134mmHg)= 36.18 mmHg

 

X(C6H5CH2CH3)= .69/(.26+.69)= .73

P(C6H5CH2CH3)= (.73)(182mmHg)= 132.86mmHg

 

57)

Ptoluene= Xtoluene*P˚toluene= (.6)(533mmHg)= 319.8 mmHg

Pbenzerne= Ptot - Ptoluene= 760-319.8=- 440.2 mmHg

 

440.2mmHg= Xbenzbenz

440.2/.4= P˚benz= 1100.5mmHg

 

59)

            ∏=MRT

           

            n/v = π/RT = ((2mmHg)(1atm/760mmHg))/((.08206)(298K))= 1.076E-4M

 

solute amount= .25 L * (1.076E-4M/1L)= 2.69E-5 mols

 

molar mass = .58g/2.69E-5 mol = 2.156E4 g/mol

 

61) The solutions inside of the plants are less concentrated than the concentrated solution outside and so the plant solution will move, or cross he membrane in order to dilute the salt solution.

 

63)

∏=(2M)(.08206Latm/molK)(293K)

∏=48 atm

 

65)

62.9 torr / 760 = .0828atm

 (assume mass of .97g)

 

.0828atm = (.97g/Molar Mass) x .0821 L*atm/mol*K x 298K

 

Molar mass= 286.7 g/mol

 

75)

m=∆Tf/-Kf=    1.3-5.48/-5.12=     .816 m

 

amount= (50 ml benz*(.879g/ml)*(1kg/1000g)*(.816 mol/kg benz)= .03586 mol

 

molec weight = 2.8867 g/ .03586 mol = 80.5011 g/ mol

 

1 mol B= 40.3g/(1.811g/mol)=3.72 mol B/3.72

1 mol N=52.2g/(14.0067 g/mol)= 3.72 mol N/3.72

2 mol H= 7.5 /(1.008 g/mol)= 7.44 mol H/3.72

 

BNH2= 26.8337                                 80/26.8337=3

 

B3N3H6

 

77)  ∆Tb=2˚C   Kbwater = .512 ˚C/m   mass H2O = 2 kg       i=2

 

m=   (∆Tb/iKb) =  2/ (2*.512)  =2 m

 

solute mass =  2.37 L watee*2 mols NaCl/kg water*58.4 g/mol NaCl =  276.816 g NaCl

 

83)

 

a)           Co(NH2)2

a.            Tf=/(1)(1.86)(.25)= -.465

b)           NH4NO3

a.            Tf= -(2)(1.86)(.25)= -.93

c)    HCl     

a.     Tf=-(2)(1.86)(.25)= -.93

d)   CaCL2

a.     Tf= -(3)(1.86)(.25)= 1.395

 

87)  

n(CO(NH2)2= .617 g / 60.06 g/mol= .0103 mol

n H20 + CO(NH2)2 = 90 g / 18.02 g/ mol = 4.99 mol

n C12H22O11= 3.7g/ 342.3  =.0108 mol

n H20 + 80 g h20 /18.02 = 4.44 mol H2O

 

.0103/.0103+nwater = XCO(NH2)2 = X C12H22O11=   .0108/(.0108+(9.85-nwater)

nwater= 4.8083

        .0103/(.0103+nwater)= XCO(NH2)2 = X C12H22O11= .00214

 

88) separation by fractional solidification.

 

113)

(58.9 g/(86g/mol)= .68 mol C6H1

44g/78 = .56 mol C6H6

 

1.24 mol tot  

XH= .44

XB= .45

Ptot = .55*573 + .43*391 = 491 mmHg

 

 

 

Q13.1

Which of the following do you expect to be most water soluble, and why? C10H8(s),NH2OH(s),C6H6(l),CaCO3(s)

Q13.2

Which of the following is moderately soluble both in water and in benzene [C6H6 (l)], and why? (a) 1-butanol, CH3(CH2)2CH2OH; (b) naphthalene, C10H8; (c) hexane, C4H14; (d) NaCl (s).

Q13.3    

Substances that dissolve in water generally do not soluble in benzene. Some substances are moderately soluble in both solvents, however. One of the following is such a substance. Which do you think it is and why?

Q13.33

Calculate the mole fraction of the solute in the following aqueous solutions: (a) 0.112M C6H12O6 (d=1.006g/ml); (b) 3.20% ethanol, by volume (d=0.993 g/ml; pure CH3CH2OH, d=0.789g/ml).

35. What volume of glycerol, CH3CH(OH)CH2OH (d=1.26g/ml), must be added per kilogram of water to produce a solution with 4.85 mol % glycerol?

39. Refer to Figure 13-8 and determine the molarity of NH4Cl in a saturated aqueous solution at 40°C.

41. A solution of 20.0kg KClO4 in 500.0 g of water is brought to a temperature of 40°C. (a) Refer to figure 13-8 and determine whether the solution is unsaturated or supersaturated at 40°C. (b) Approximately what mass of KClO4, in grams, must be added to saturated the solution (if originally)

43. Under an O2 (g) pressure of 10.00 atm, 28.31 ml of O2 (g) dissolves in 1.00 L H2O at 25°C. What will be the molarity of O2 in the O2 in the saturated solution at 25°C when O2 pressure is 3.86 atm? (Assume that the solution volume remains at 1.00L)?

45. Natural gas consists of about 90% methane, CH4. Assume that the solubility of natural gas at 20°C and 1 atm gas pressure is about the same as that of CH4, 0.02g/kg water. If a sample of natural gas under a pressure of 20 atm is kept in contact with 1.00*103 kg of water, what mass of natural gas will dissolve?

47. The aqueous solubility at 20°C of Ar at 1 atm is equivalent to 33.7 ml Ar (g), measured at STP, per liter of water. What is the molarity of Ar in water that is saturated with air at 1 atm and 20°C? Air contains 0.934% Ar by volume. Assume that the volume of water does not change when it becomes saturated with air.

49. Henry s law can be stated this way: The mass of a gas dissolved by a given quantity of solvent at a fixed temperature is directly proportional to the pressure of the gas. Show how this statement is related to equation (13.2).

51. What are the partial and total vapor pressures of a solution obtained by mixing 35.8 g benzene, C6H6, and 56.7 g toluene, C6H5CH3, at 25 °C? At 25 °C, the vapor pressure of C6H6 = 95.1 mmHg; the vapor pressure of C6H5CH3 = 28.4 mmHg.

53. Calculate the vapor pressure at 25 °C of a solution containing 165 g of the nonvolatile solute, glucose, C6H12O6, in 685 g H2O. The vapor pressure of water at 25 °C is 23.8 mmHg.

55. Styrene, used in the manufacture of polystyrene plastics, is made by the extraction of hydrogen atoms from ethylbenzene. The product obtained contains about 38% styrene (C6H5CH = CH2) and 62% ethylbenzene (C6H5CH2CH3), by mass. The mixture is separated by fractional distillation at 90 °C. Determine the composition of the vapor in equilibrium with this 38% 62% mixture at 90 °C. The vapor pressure of ethylbenzeneis 182 mmHg and that of styrene is 134 mmHg.

57. A benzene-toluene solution with xbenz = 0.300 has a normal boiling point of 98.6 °C. The vapor pressure of pure toluene at 98.6 °C is 533 mmHg. What must be the vapor pressure of pure benzene at 98.6 °C? (Assume ideal solution behavior.)

59. A 0.72 g sample of polyvinyl chloride (PVC) is dissolved in 250.0 mL of a suitable solvent at 25 °C. The solution has an osmotic pressure of 1.67 mmHg. What is the molar mass of the PVC?

61. When the stems of cut flowers are held in concentrated NaCl (aq), the flowers wilt. In a similar solution a fresh cucumber shrivels up (becomes pickled). Explain the basis of these phenomena.

63. In what volume of water must 1 mol of a nonelectrolyte be dissolved if the solution is to have an osmotic pressure of 1 atm at 273 K? Which of the gas laws does this result resemble?

65. At 25 °C a 0.50 g sample of polyisobutylene (a polymer used in synthetic rubber) in 100.0 mL of benzene solution has an osmotic pressure that supports a 5.1 mm column of solution (d = 0.88 g/mL). What is the molar mass of the polyisobutylene? (For Hg, d=13.6 g/ml.)

75. Thiophene (fp = - 38.3; bp = 84.4 °C) is a sulfur containing hydrocarbon sometimes used as a solvent in place of benzene. Combustion of a 2.348 g sample of thiophene produces 4.913 g CO2, 1.005 g H2O, and 1.788 g SO2. When a 0.867 g sample of thiophene is dissolved in 44.56 g of benzene (C6H6), the freezing point is lowered by 1.183 °C. What is the molecular formula of thiophene?

77. Cooks often add some salt to water before boiling it. Some people say this helps the cooking process by raising the boiling point of the water. Others say not enough salt is usually added to make any noticeable difference. Approximately how many grams of NaCl must be added to a liter of water at 1 atm pressure to raise the boiling point by 2 °C? Is this a typical amount of salt that you might add to cooking water?

83. NH3 (aq) conducts electric current only weakly. The same is true for acetic acid, HC2H3O2(aq).When these solutions are mixed, however, the resulting solution conducts electric current very well. Propose an explanation.

87. A typical root beer contains 0.13% of a 75% solution by mass. How many milligrams of phosphorus are contained in a 12oz can of this root beer? Assume a solution density of 1.00 g/mL; also, 1oz = 29.6mL.

88. An aqueous solution has 109.2 g KOH/L solution. The solution density is 1.09 g/ml. Your task is to use 100.0 mL of this solution to prepare 0.250 m KOH. What mass of which component, KOH or H2O, would you add to the 100.0 mL of solution?

103. Instructions on a container of antifreeze (ethyleneglycol; fp, - 12.6 °C, bp, 197.3 °C) give the following volumes of Prestone to be used in protecting a 12 qt cooling system against freeze-up at different temperatures (the remaining liquid is water): 10 °F, 3 qt; 0 °F, 4 qt; - 15 °F, 5 qt; - 34 °F, 6 qt. Since the freezing point of the coolant is successively lowered by using more antifreeze, why not use even more than 6 qt of antifreeze (and proportionately less water) to ensure the maximum protection against freezing?

113. Cinnamaldehyde is the chief constituent of cinnamon oil, which is obtained from the twigs and leaves of cinnamon trees grown in tropical regions. Cinnamon oil is used in the manufacture of food flavorings, perfumes, and cosmetics. The normal boiling point of cinnamaldehyde, C6H5CH = CHCHO, is 246.0 °C, but at this temperature it begins to decompose. As a result, cinnamaldehyde cannot be easily purified by ordinary distillation. A method that can be used instead is steam distillation. A heterogeneous mixture of cinnamaldehyde and water is heated until the sum of the vapor pressures of the two liquids is equal to barometric pressure. At this point, the temperature remains constant as the liquids vaporize. The mixed vapor condenses to produce two immiscible liquids; one liquid is essentially pure water and the other, pure cinnamaldehyde. The following vapor pressures of cinnamaldehyde are given: 1 mmHg at 76.1 °C; 5 mmHg at 105.8 °C; and 10 mmHg at 120.0 °C. Vapor pressures of water are given in Table 13.2.(a) What is the approximate temperature at which the steam distillation occurs?(b) The proportions of the two liquids condensed from the vapor is independent of the composition of the boiling mixture, as long as both liquids are pre- sent in the boiling mixture. Explain why this is so. (c) Which of the two liquids, water or cinnamaldehyde, condenses in the greater quantity, by mass? Explain.

117. In your own words, define or explain the following terms or symbols: (a) xB; (b) PA°; (c) Kf; (d) i; (e) activity.

 

Answers

1.      NH2(OH) (s)

2.      NaCl. The attractions between unlike molecules are much weaker and the components remain segregated on a heterogeneous mixture.

3.      C

33. (a). 0.00204 (b) 0.0101

35. 207ml

39. 8.66m

41. (a) Unsaturated. (b) 3g

43. 4.47*10-3M

45. 400g

47. 1.4*10-5M

49. Because of the low density of molecules in gaseous state, the solution volume remains essentially constant as a gas dissolves in liquid.

51. 56.9mmHg

53. 23.2mmHg

55. 0.32

57. 1290mmHg

59. 3.2*104g/mol

61. Both two are less concentrated then the salt solution.

63. 22.4L

65. 2.8*105g/mol

75. C4H4S

77. 120g

83. NH3(aq) +HC2H3O(aq)àNH4C2H3O2(aq)

      NH4C2H3O2(aq) àNH4+(aq)+C2H3O2-(aq)

87. 0.462g

88. 682g H2O

103. 113

117. (a) Mole fraction

        (b) Vapor pressure

        (c) Depression constant

        (d) Van’s Hoff introduced factor

 

 

 

 

 

3.) Substances that dissolve in water generally do not dissolve in benzene. Some substances are moderately soluble in both solvents, however. One of the following is such a substance. Which do you think it is and why?

                  Salicyl Alcohol                                  Hydrochloric Acid                                             Oxyacetic Acid                  

 

Answer: Salicyl Alcohol because of its OH groups and the benzene ring

http://chemwiki.ucdavis.edu/Analytical_Chemistry/Chemical_Reactions/Properties_of_Matter/Solubility_Rules

 

43.) Under an O2(g) pressure of 1.00 atm, .03522 L of O2(g) dissolves in 1 L of water at 25°C. What will be the molarity of O2 in the saturated solution at 25°C when the O2 pressure is 4.88 atm? (Assume that the solution volume remains at 1 L).

Answer: n = PV/RT =( 1)(.03522)/(0.0821)(298) = .0014 mol

So molarity = .0014 mol/1.0 L = .0014 M

C = KP, so K = C/P

k = .0014 M/ 1 atm

When O2 pressure is 4.88 atm…

.0014 = C/(4.88)

C (concentration of O2) = .0068 M

http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Solutions/Solubilty/Types_of_Saturation

 

53.) Calculate the vapor pressure at 25°C of a solution containing 200 g of the nonvolatile solute, glucose, C6H12O6, in 700 g of water. The vapor pressure of water at 25°C is 23.8 mmHg.

Answer: Raoult’s Law: PA = (XA)(P°A)

200 g C6H12O6 = 1.11 mol

700 g H2O = 38.89 mol

Mole fraction (XA) = (38.89)/(38.89 + 1.11) = .0278 mol

P°A = 23.8 mmHg (vapor pressure of the pure solvent at 25°C)

So, PA = (.0278)(23.8) = .6605 mmHg

http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Solutions/Ideal_Solutions/Changes_In_Vapor_Pressure%2c_Raoult's_Law

 

63.) In what volume of water must 1 mol of a nonelectrolyte be dissolved if the solution is to have an osmotic pressure of 3.0 atm at 298 K? What exactly is osmotic pressure?

Answer: Use equation π =(n/V)RT

3 = (1/V)(.0821)(298 K)

3 = (1/V)(24.47)

V = 8.16 L

Osmotic pressure is the necessary pressure required to stop osmotic flow (net flow of water) in a solution.

http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Solutions/Colligative_Properties/Osmotic_Pressure

Q13.87

A soda contains 0.15% of an 80% CO2 solution by mass. How many milligrams of carbon are contained in a 12 oz can of soda? Assume a solution density of 1.00 g/mL and 1 oz = 29.6 mL.

S13.87

12 oz = 355.2 mL

\[\rho = \dfrac{M}{V}\]

1 = M/355.2 mL

M = 355.2 grams x.80 = 284.16 grams CO2

0.15% = grams of C/284.16 grams CO2 x100

= 0.426 grams = 462 miligrams of carbon

http://chemwiki.ucdavis.edu/Analytical_Chemistry/Quantifying_Nature/Density_and_Percent_Compositions

 

 

 

 

 

 

 

1. Which of the following is to be expected to be the most soluble in hexane and why? CH3OH, NaI, C5H12

Answer:

C5H12 ,which is pentane, is non polar just like hexane and components with like properties are more soluble in each other because they are similar. Both are non polar because of the small difference in electronegativity between Carbon (+4) and Hydrogen (+1).

 

2. Which of the following is moderately soluble both in water and in benzene [C6H6], and why? (a) Phenol, C6H5OH (b) Methane, CH4 (c) Hexane, C6H14 (d) Oxygen, O2

Answer:

(a) Phenol, C6H5OH is the only polar compound, therefore the only one soluble in water and benzene.

 

# 3 – Why would Salicyl alchohol dissolve in both water and benzene while most other substances don’t?

                Salicyl alcohol contains a ring of benzene and can use its –OH groups to hydrogen bond to water molecules

 

# 33 – Calculate the amount of solvent in a 0.1M Fe4Cl12 solution with density 3.45 g/mL

solvent amount = (( 1 L soln * (1000ml/1L) * (3.45g/1mol) - (.1molFe4Cl12 * (501.4g/1mol))

= 3399.86 * (1 mol H20/18.02g H20) = 188.7 mol H20

 

35.) What volume of glucose, C6H12O6 (d=1.54 g/ml), must be added per kilogram of water to produce a solution with 3.23 mol % glucose?

Answer:

3.23% = (x)(100)       x=0.0323  moles

0.0323mol (180.18g/1mol)= 5.819814g glucose     D=M/V      V=M/D  V= 5.819814/1.54       V=3.78ml

 

#39 – At 30C is 13.2g per 100grams of water. Calculate the molarity.

molarity = (13.2g * (1mol K2SO4 / 174.6g K2SO4))/(100gH20 * (1kg/1000g)) = .756 m

 

41.) A solution of 32.0 g KNO3 in 11.0 g of water is brought to a temperature of 35°C.

(a) Refer to Figure 13-8 and determine whether the solution is unsaturated or supersaturated at 35°C.

Answer:  The solution is unsaturated at 35°C because on the KNO3 curve 32g and 35°C is in the region under the curve.

(b) Approximately what mass of KNO3, in grams, must be added to saturate the solution (if originally unsaturated), or what mass of KNO3 can be crystalized (if originally supersaturated)?

Answer:

11.0g H2O x (53g KNO3/100g H2O)= 5.83g

32g-5.83g=26.17g

 

43.) Under an N2 (g) pressure of 1.0atm, 21.41 ml of N2(g) dissolves in 3.00L H2O at 0°C. What will be the molarity of N2 in the saturated solution at 0°C when the N2 pressure is 5.83 atm? (Assume that the solution volume remains at 3.00L).

Answer:  

Molarity= [0.0214L N2 x (1 mol N2/22.4L N2 (STP))]/1L soln = 9.558 x 10-4 M N2

K=C/Pgas =9.558 x 10-4 M N2/1.00atm

C=k x Pgas = (9.558 x 10-4 M N2/1.00atm) x 5.83atm= 0.005572 M N2

 

45.       If the solubility of H2 gas is .015 g / atm • kg H2O, calculate the mass of H2 gas that dissolves in a vessel containing hydrogen gas at 4.3 atm and 254 kg of water.

 

            Mass H2 dissolves = (4.3 atm)(254 kg water)(.015 g / atm • kg water) = 16.4 g H2

 

47.       a) A closed container contains of 5.40 L of liquid water and Ne gas at 2.45 atm. The quantity of neon gas dissolved in the water is equivalent to 28.9 grams of Ne. Calculate the solubility of Ne in water in moles Ne / kg H2O • atm).

 

            mol = (28.9 g Ne)(1 mol / 20.18 g) = 1.43 mol Ne

            mass water = (5.40 L water)(1000 mL / 1 L)(1 mL / 1 cm^3)(1 g / 1 cm^3)(1 kg / 1000 g) =

                        =5.40 kg H2O

            solubility Ne = (1.43 mol Ne) / (5.40 kg water)(2.45 atm) = 0.108 mol / kg H2O • atm

 

            b) What is the molarity of the dissolved neon?

 

            Molarity = mol solute / Liter solvent = (1.43 mol Ne) / (5.40 L water) = 0.265 M

 

49.       4.52 g of an unknown compound reduces the freezing point of 65.14 g of ethanol (Kf = 1.99 K • kg / mol) from -114 °C to -145 °C. What is the molar mass of this substance?

            ∆Tf = -31 °C = -31 K                                                 ∆Tf = -iKfm

            -31 K = -2(1.99)(mol / .06514 kg)                  mol = .507 mol substance

            molar mass = 4.52 g / .507 mol = 8.91 g / mol

51) if the partial pressure for gas A is 44.11 mmHg and for gas B is 22.18 mmHg what is the total pressure of a mixture of 1 mole of each?

 

Ptot = P1 + P2… 44.11 mmHg + 22.18 mmHg = 66.29 mmHg

 

53) Calculate the vapor pressure at 25C° of solution containing 180g of C6H12O6, in 700g H2O). The vapor pressure of the water at 25C is 24.8 mmHg.

 

First solve for mole fraction of water, 200g C6H12O6 is 1 mole, 700 grams water is 38.8 moles. (Moles water)/total moles = 38.8/39.8 = .974

P = Xwater x Pwater = 24.8 mmHg * (.974) = 24.15 mmHg

 

57: Assume that the vapor pressure of water is 0.40 atm at 25 degrees Celsius. What is the vapor pressure of a solution of 100 grams water and 50 grams of C6H12O6? Use  Raoult's law.

Answer:

PH2O = XH2O * P*H20

XH2O = (100 g H2O * 1mol/18g) / ((100 g H2O * 1mol/18g) + (50 g C6H12O6*1mol/180g)) = 0.95

PH2O = XH2O * P*H20 = 0.95 * 0.4 atm = 0.38 atm

# 59 – A 0.89g sample is dissolved in 250mL of solvent at 25C with osmotic pressure 2.39mmHg. What is the molar mass of the sample?

(n/V) = (p/RT) = (2.39mmHg * (1atm/760mmHg)/(0.08206 L atm mol^-1 K^-1 * 298.2K = 1.29*10^-4 M

solute amount = 0.25L * (1.29*10^-4)/(1L) = 3.21*10^-5 mol             M = (.89g)/(3.21*10^-5) = 2.8*10^4g/mol

 

61.) What is the process of osmosis?

 Osmosis is the process where water travels from a high concentration to a low concentration through a semi-permeable membrane.

 

63.       How many moles of a nonelectrolyte must be dissolved in 2.30 L of water to form a solution that has an osmotic pressure of 4.30 atm at 298 K?

 

            p = MRT                    4.30 = (mol / 2.30 L)(.08206)(298 K)

                                                mol = .404 mol

65) What are the factors that contribute to osmotic pressure?

Van’t Hoff factor = i

Temperature = T (in kelvin)

Molarity = M

Universal gas constant = R

 

75. The molecular formula of thiopene is C4H4S.  How many grams of CO2, H2O, and SO2 are produced when 0.867 grams of thiopene are combusted?

Answer:

MW thiopene = 4*12.01 + 4*1.008 + 32.065 = 84.137

Combustion of thiopene:

C4H4S + 6O2 -> 4CO2 + 2H2O + SO2

-> moles thiopene = 0.867 [gms] / 84.137 [gms/mole] = 0.0103 moles

-> gms CO2 = 4 * 0.0103 moles * 44 gms/mole = 0.453 gms

-> gms H2O = 2 * 0.0103 moles * 18 gms/mole = 0.185 gms

-> gms SO2 = 1 * 0.0103 moles * 32 gms/mole = 0.329 gms

 

# 77 – How much NaOH is required to change the boiling point of water by 5C?

m = (delta T)/(i * K) = (5C)/(2.00 * 0.512C/m) = 4.9m

solute mass = 1L H2O * (1kg H20/1L H2O) * (2mol NaCl/1kg H2O) * (58.4g NaCl/mol NaCl) = 286.16g NaCl

 

83.) How can you create a solution that is a good conductor of electric current?

Answer:  You can create a solution that is a good conductor by mixing two weak conductors, one a weak acid and one a weak base, that form a salt solution and water. By having a weak acid and a weak base it means that there are fewer ions present. When the two solutions are combined, ions are present, which means that the conductivity increases which means that a strong electrolyte forms.  

 

87.       A 16 oz bottle of beer is 4.5% alcohol by volume. Calculate the mass of this ethanol alcohol if it has a density of .789 g/mL. (1 oz = 29.6 mL).

 

            mL = (16 oz)(29.6 mL / 1 oz) = 473.6 mL

            volume ethanol = (473.6 mL)(.045) = 21 mL

            mass ethanol = (21 mL)(.789 g / mL) = 16.8 g ethanol

Q13.88

An aqueous solution has 1 M KOH. Use 200 ml of this solution to prepare .250 M KOH what mass of which component, KOH or H2O would you add to the 200 ml of solution.

S13.88

You must add H2O to lower molar concentration. .2 moles exist in the solution, therefore

0.2/ 200 ml + x ml H2O = 0.25 M

X = 600 ml H2O

 

# 113 – A solution of KI contains 256g KI per every 100g water. What is the percent mass of the KI?

%KI = (256g KI) / (256g KI + 100g H20) * 100% = 71.9% = 71.9g KI/ 100g solution

Q13.117

Define or explain the following terms (a) conductivity (b) osomotic pressure (c) Molarity (d) supersaturated solution (e) molality  

S13.117

  1. Saturated solution: It is when the quantity of dissolved solute stays constant with time.  
  2. Osmotic Pressure: It is the necessary pressure to stop the osmotic flow of a solution. The equation for solving osmotic pressures of dilute solutions of nonelectrolytes is p=M X RT.
  3. Molarity: It is the conversion factor which relates amount of solute to the volume of solution.
  4. Supersaturated Solution: It is when the amount of solute is greater than the quantity in a saturated solution and supersaturated solutions are unstable.
  5. Molality: It is the amount of solute (moles) divided by the mass of solvent (in Kg)

Q13.103

Why not use 100% pure ethylene glycol as an antifreeze:

S13.103

Because the freeze point is at a minimum at approximately 50% ethylene glycol / water therefore the pure ethylene glycol will not make a suitable antifreeze.

 

 

Arrange in order of most soluble to least. Explain why you have arranged them in order.
 
Na_2 CO_3 (s),   C_2 H_4 (g),  CH_3 (l),  CaCl_2 (s)
  
 
CaCl_2 (s) the most soluble
Na_2 CO_3 (s), only slightly soluble because of the carbonate
C_2 H_4 (g),  CH_3 (l), nonpolar molecules are insoluble in water.
 
 
Some substances are only soluble in water. Some substances are only soluble in acid. Which of the following is soluble in acid and water?
 
Al(ClO_3 )_3,   Al(OH)_3,   Al_2 SiO_5,   Al_2 O_3
 
Al(OH)_3 soluble in acids, insoluble in water
Al_2 O_3 slightly soluble in acids, insoluble in water
     Al_2 SiO_5  insoluble in acids and water
     Al(ClO_3 )_3 soluble in water
 
 
Which of the following would be soluble in HCl but not in water?
 
Ag_3 PO_(4,),   AgBr,   AgI,   AgClO_3
 
Ag_3 PO_(4,) soluble in acid, insoluble in water
AgBr slightly in HCl, insoluble in water
AgI, insoluble
AgClO_3 soluble in water
 
 
33. The density of a solution of 55.5g CaCl_2  (MM=110.98 g/mol) and 500mL of water is 1.19 g/mL. Calculate the mole fraction of water.
 
55.5g CaCl_2  ((1 mol CaCl_2)/(110.98gCaCl_2 ))=.5 mol
 
500g H_2 O (K/(10^3 ))=.5 kg
 
=(.5 mol CaCl_2)/(.5 kg H_2 O)=1m
 
500g H_2 O+55.5g CaCl_2=55.5 g solution
 
555.5g solution ((1×〖10〗^(-3) L)/(1.19 g solution))=(.5 mol CaCl_2)/(.4668 L solution)=1.07M
X_(H_2 O)=n_(H_2 O)/(n_(H_2 O)+n_(Ca^(2+) )+ n_(Cl^- ) )
 
X_(H_2 O)=4.5/(4.5+.5+1)=.7509
 
35. The density of Ethylene glycol (C2H6O2) is 1.11g/mL. To produce a solution with 3.0 mol % C2H6O2, what volume must be added per kg of water?
 
 
n_water=1000 g H_2 O ×  (1 mol H_2 O)/(18.02 g H_2 O)=55.49 mol H_2 O
 
n_( C_2 H_6 O_2 )=3.0%=0.030 mol C_2 H_6 O_2
 
X_( C_2 H_6 O_2 )=0.030=n_( C_2 H_6 O_2 )/(n_( C_2 H_6 O_2 )+55.49)
 
n_( C_2 H_6 O_2 )=0.030n_( C_2 H_6 O_2 )+1.6647
 
n_( C_2 H_6 O_2 )=1.6647/((1-0.030))=1.716 mol C_2 H_6 O_2
 
1.716 mol C_2 H_6 O_2×(62.07 g C_2 H_6 O_2)/(1 mol〖 C〗_2 H_6 O_2 )×(1 mL)/(1.11 g)=95.96 mL C_2 H_6 O_2
 
 
 
 
41) A solution of 30.0 grams K_2 〖SO〗_4 in 400 grams of water is brought to a temperature of 20°C given that at 20°C, a saturated K_2 〖SO〗_4 solution has a concentration of about 12 grams K_2 〖SO〗_4 dissolved in 100 grams of water.
a) Is the solution unsaturated or supersaturated? 
b) Approximately what mass of KNO3, in grams, must be added to the solution (if originally unsaturated) or what mass of KNO3 can be crystallized (if originally supersaturated)
 
 
a) 
(Mass solute)/(100 g H_2 O)=100 g H_2 O*  (30 g K_2 〖SO〗_4  )/(400 g H_2 O)  =7.5 grams K_2 〖SO〗_4
 
→ The solution is thus unsaturated. 
 
b) 
(400 grams H_2 O* (12 g〖 K〗_2 〖SO〗_4  )/(100 g H_2 O))- 30.0 g K_2 〖SO〗_4=18 g 〖 K〗_2 〖SO〗_4
 
 
 
45) Assume that the solubility of a natural gas at 20°C and 1 atm gas pressure is .03 g/kg of water. If a sample of natural gas under a pressure of 15 atm is kept in contact with 〖2.00×10〗^3  kg of water, what mass of natural gas will dissolve? 
 
Mass of natural gas = 〖2.00×10〗^3  kg * (.03 g natural gas)/(1 kg H_2 O atm) * 15 atm =
 
= 〖9×10〗^2 g natural gas 
 
47) The aqueous solubility at 25°C of 〖CO〗_2  (g) at 1 atm is equal to 41.6 mL 〖CO〗_2  (g), measured at standard temperature and pressure, per liter of water. What is the molarity of 〖CO〗_2  (g) in water that is saturated with air at 25°C and 1 atm? Air contains .039% 〖CO〗_2 by volume. 
 
K_(〖CO〗_2 )   = C/P_(〖CO〗_2 )  = (((41.6 mL 〖CO〗_2)/(1 L solution)* (1 mol 〖CO〗_2)/(22,400 mL at STP))/(1 atm) = .(00186 M)/(1 atm)
 
Partial pressure 〖CO〗_2 = .000039 atm 
 
C = K_(〖CO〗_2 ) P_(〖CO〗_2 ) = (.00185 M)/(1 atm) * .000039 atm = 47.69 M 
 
49) Explain why the volume of a gaseous solution remains essentially constant as a gas dissolves in a liquid. Also what equation does this help to explain? 
 
In a gaseous state, the molecules in the solution have a very low density. The changes in the concentration of the solution are not directly proportional to the volume of the solution, but rather the number of dissolved gas molecules. The mass of the gas dissolved is proportional to the pressure of the gas itself. This statement helps to explain the equation C=K × P_gas, also known as Henry’s Law. 
 
51) What are the partial and total vapor pressures of a solution obtained by mixing 41.9 g methane, 〖CH〗_4, and 62.3 g ethanol, C_2 H_6 O at 30°C? At 30°C the vapor pressure of 〖CH〗_4 is 51.2 mmHg; the vapor pressure of C_2 H_6 O is 31.8 mmHg. 
 
n_M=41.9 g 〖CH〗_4 * (1 mol 〖CH〗_4  )/(16.04 g 〖CH〗_4  ) = 2.61 mol 〖CH〗_4
 
n_E=62.3 g C_2 H_6 O * (1 mol C_2 H_6 O )/(46.07 g C_2 H_6 O ) = 1.35 mol C_2 H_6 O
 
X_M = (2.61 mol 〖CH〗_4)/(2.61+1.35) = .659 mol 〖CH〗_4
 
X_E = (1.35 mol C_2 H_6 O)/(2.61+1.35) = .341 mol C_2 H_6 O
 
P_M = 2.61 mol 〖CH〗_4 * (51.2 mmHg)/(1 mol 〖CH〗_4 ) = 133.6 mmHg 〖CH〗_4
 
P_E = 1.35 mol C_2 H_6 O * (31.8 mmHg C_2 H_6 O)/(1 mol C_2 H_6 O) = 42.93 mmHg C_2 H_6 O
 
P_total=133.6+42.93=176.53 mmHg
 
53) Calculate the vapor pressure at 30°C of a solution containing 175 g of the nonvolatile solute, salt, NaCl, in 725 g H_2 O. The vapor pressure of water at 30°C is 31.8 mmHg. 
 
 
n_salt = 175 g NaCl * (1 mol NaCl)/(58.44 g NaCl) = 2.99 mol NaCl
 
n_water = 725 g H_2 O * (1 mol H_2 O)/(18.02 g H_2 O) = 40.23 mol H_2 O
 
x_water = (40.23 mol H_2 O)/((40.23+2.99)) = .931 mol 
 
P_solution = X_water P_water = .931 * 31.8 mmHg = 29.6 mmHg
 
43) Under an N2(g) pressure of 1.00 atm, 23.54 ml of N2(g) dissolves in 1.00 L H2O at 0 degrees celcius. What will be the molarity of N2 in the saturated solution at 0 degrees celcius when the N2 pressure is 2.00 atm? (solution volume is still 1 L.)
Solution:
To calculate the new molarity we need to find Henry’s law constant for the gas and multiply it by the new pressure.
pV/rt=n
(atm X .02354)/(.08206 L atm mol^(-1) K^(-1)  X 273K)= 1.0578X10-3 Mol N2= 1.0578X10-3 M
[N2]= (1.0578X10^-3)/(1.00 atm) X 2.00 atm= 2.1X10-3
55) A mixture is composed of 54%  H2O(l) and 46% CO2(l). Determine the vapor pressure ov each liquid when the two portions are separated at 105 degrees celcius. Assume the vapor pressure at 105 degrees celcius of CO2=120 mmHg and the vapor pressure of H2O = 105 mmHg. (these are not the actual values)
Solution: First find the mole fractions of each compound, and then multiply them by their respective pressures.
Assume 100 grams of the solution, so 54 grams of H2O and 46 grams of CO2.
Moles H2O=54 g/18 g/mol=3 mol
Moles CO2=46g/44 g/mol=1.045 mol
Vap. Press. H2O= (3/(1.045+3))(105 mmHg)=77.874 mmHg
Vap. Press. CO2= (1.045/4.045)(120 mmHg)= 31.001 mmHg
57) A solution composed of Element A and Element B have their normal boiling point at 100 degrees celcius. If xA=.400 and the vapor pressure of pure element B at 100 degrees celcius is 320 mmHg, what is the vapor pressure of pure element A at 100 degrees celcius? (assume ideal solution behavior)
Solution:
If xA=.4, xB=.6
Partial vapor pressure of element B= .6X320= 192 mmHg.
Since one assumes ideal solution behavior, the total vapor pressure at the solution’s normal boiling point is 760 mmHg.
So partial pressure of Element A= 760-192= 568 mmHg
568 mmHg=.4(pure pressure Element A)
Vapor pressure of pure element A=1412.5 mmHg
59) A 1.5 gram sample of a uknown element is dissolved in 500 ml of a suitable solvent at 37 degrees celcius. The solution has an osmotic pressure of 2.00 mmHg. What is the molar mass of the uknown element?
Solution:
Simply plug in the values to equation 13.4
N=πV/RT=(2X.5)/.08206X300= .04062 moles 
Molar mass of uknown element= 1.5 g/.04062 mol= 36.9276 g/mol
61) When organisms consume large amount of salty food at one time, they tend to get thirsty for water. Why does this occur?
Solution: When one consumes large amounts of salt, it causes a chemical imbalance outside and inside of bodily cells. To dilute the extreme salt concentration outside of the bodily cells, the low salt concentration solution that is present in the cells travels across the semi-permeable membrane to dilute the high salt concentration outside the cells. This lowers the water concentration inside bodily cells, triggering a thirst for water to replenish the cells water supplies.
 
63) In what volume of water must 5 mol of a nonelectrolyte be dissolved if the solution is suppose to have an osmotic pressure of 2 atm at 390K? Which gas laws resemble this result?
 
n/V=π/RT=(2.00 atm)/((0.08206 atm)/(mol K)× 390 K)=0.0625 M
 
Volume=5 mol ×(1 L)/(0.0625 mol solute)=80 L solution ≈80 L solvent
 
We have assumed that the solution is so dilute that its volume is basically the same volume as the solvent constituting it. This volume corresponds to the STP molar volume of an ideal. The equation for osmotic pressure also closely resembles the ideal gas equation. 
 
65) At 22°C a 0.40 g sample of polypropylene (a polymer used in textiles) in 85.0 mL of benzene solution has an osmotic pressure that supports a 6.6 mm column of solution (d = 0.75 g/mL). What is polypropylene’s molar mass? (For Hg, d = 13.6 g/mL).
 
Determine the concentration of the solution from the osmotic pressure
 
π= 6.6 mm soln.  ×(0.75 mmHg)/(13.6 mm soln.)  ×  (1 atm)/(760 mmHg)=4.8 ×〖10〗^(-4)  atm
n/V=  π/RT=  (4.8 ×〖10〗^(-4)  atm)/((0.08206 L atm)/(mol K)  × 295K )=2.0 ×〖10〗^(-5)  M
Amount of Solute
=85.0 mL ×(1 L)/(1000 mL)  × 2.0 ×〖10〗^(-5)  M=1.7 ×〖10〗^(-6) mol solute
Molar Mass
=  (0.40 g)/(1.7 × 〖10〗^(-6)  mol)=(〖2.4 ×10〗^5 g)/mol
 
75) Benzene  (fp= 5.5; bp = 80.1°C) is a hydrocarbon, thiophene is sometimes used as a solvent in its absence. Combustion of a 2.543 g sample of benzene produces 5.150 g H2O, and 1.699 g of CO2. The freezing point is lowered by 0.893°C, when a 0.782 g sample of benzene is dissolved in 39.72 g of thiophene (C4H4S). What is the molecular formula of benzene? (thiophene; Kf =(4.72°C)/m)
 
First, determine the molality of the thiophene solution, then the molar mass of the solute. 
 
m=〖∆T〗_f/〖-K〗_f =(-0.893°C)/((-4.72°C)/m)=0.189m
amount of solute=0.03972 kg thiophene×(0.189 mol solute)/(1 kg thiophene)=0.00751 mol solute
 
molar mass=  (0.782 g benzene)/(0.00751 mol benzene)=(104.1 g)/mol
Next, use the masses of the combustion products to determine the empirical formula.
 
Amount of C= 1.699 g 〖CO〗_2×(1 mol 〖CO〗_2)/(44.010 〖CO〗_2 )×(1 mol C)/(1 mol 〖CO〗_2 )=0.0386 mol C ÷0.0386=1 mol C
 
Amount of H=5.150g ×(1 mol H_2 O)/(18.015 H_2 O)×(2 mol H)/(1 mol H_2 O)=0.572 mol H÷0.0386=14 mol H
This gives us a molecular formula of CH14, which for benzene is incorrect. Although the math was correct, the estimates given on how much H2O and CO2 benzene produces due to combustion were wrong. 

Q13.77

Many people when boiling water add salt, believing that it helps raise the boiling point, shortening the cooking process. About how many grams of NaCl would you need to add to 5.0 L of water at 1 atm pressure to raise the boiling point by 5°C? Is this more than a person would typically add to their cooking water?
 
m=(∆T_b)/(i×K_b )=(5°C)/(2.00 × (0.512°C)/m)=4.88 m
 
solute mass=5.0 L H_2 O ×  (1 kg〖 H〗_2 O )/(1 L H_2 O)  ×  (2 mol NaCl)/(1 kg H_2 O)  ×  (58.4 g NaCl)/(1 mol NaCl)=1460g NaCl
This is at least two hundred times the amount of a salt a person would normally when adding salt for cooking purposes.  

Q13.83

HOC6H5(aq) is a poor conductor of electricity. The same is to be said about ammonia, NHs(aq). However, when mixed, these resulting solutions conduct electricity very well. Offer an explanation. 
 
The combination of HOC6H5(aq) with NHs(aq), results in the formation of NH4OC6H5(aq), which is a solution of NH4+ and C6H5O- ions. 〖NH〗_3 (aq)+ 〖HC〗_2 H_3 O_2 (aq)→〖NH〗_4 〖OC〗_6 H_5 (aq)→〖〖NH〗_4〗^+ (aq)+〖C_2 H_3 O_2〗^- (aq). This solution of ions or strong electrolytes conduct electricity very well. 
 
 

Q13.87

A typical can of soda may contain 0.11% of an 85% H_3 PO_4 solution by mass. How many miligrams of phosphorus are contained in a half-liter bottle (about 16.9 ounces) of soda? You may assume the solution density to be 1.00 g/ml and that 1 ounce = 29.6 ml. 
 
Start by solving for the mass of the solution
Density = mass⁄volume    Mass = Density * Volume
Mass of solution (soda)= 1 g⁄ml  × (16.9 ounces × (29.6 ml)/(1 ounce))=500.24 mass soda
Of the total solution,0.11% of the mass is the H_3 PO_4 solution.  
500.24g × 0.0011=0.55 grams H_3 PO_4  solution. 
In the H_3 PO_4  solution,85% is H_3 PO_4
0.55 g H_3 PO_4  solution × 0.85=0.468 g H_3 PO_4
For every 98 grams of H_3 PO_4,there are 30.97 grams of P
0.468g H_3 PO_4×(30.97 g P)/(98g H_3 PO_4 )  ×  (1,000 mg)/(1 gram)=147.9 mg of P per half liter of soda

Q13.88

A solution has 120.8 g of NaOH/Liter solution, with a solution density of 1.21. Use 50 ml of this solution to prepare 0.1 M NaOH. What mass of which component, NaOH or H_2O would you add to the 50ml solution?

S13.88

First, solve for the molarity of the original solution
 
NaOH molatiry
\[(120.8 g NaOH × ((1 mol NaOH)/(40 g NaOH)))/(1 L of solution)=3.02 M NaOH\]
This solution is MORE concentrated than the desired 0.1 M solution. Thus, this solution needs to be diluted. Determine the mass of water produced in the final solution, and the mas of water in the original solution, and finally the mass of water that needs to be added to dilute the solution. 
 
Mass of H_2 O in final solution=0.05 L original solution ×  (3.02 M NaOH)/(1 L solution)  ×  (1 kg H_2 O)/(0.1 M NaOH)=1.51 kg H_2 O
Mass of the original solution=50 ml ×(1.21 g)/(1 ml)=60.5 grams original solution
Mass of NaOH=50 ml ×  (1 L)/(1000 ml)  ×  (60.5 g NaOH)/(1 L solution)=3.025 g NaOH
 
Original mass of water = 60.5g solution-3.025g NaOH=57.5g H_2 O
Mass added of H_2 O=1510g H_2 O-57.5g H_2 O=1452.5g H_2