13: Solutions

Prelude

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Exercise \(\PageIndex{1a}\)

_____ prevents the dissolution of water in octane (C₈H₁₈).

1. Dipole-dipole attraction between octane molecules
2. Hydrogen bonding between molecules
3. Ion-dipole attraction between water and octane molecules
4. London dispersion forces between octane molecules
5. Repulsion between like-charged water and octane molecules

Answer

b. Hydrogen bonding between molecules

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Exercise \(\PageIndex{1b}\)

Which of the following cannot constitute a solution?

1. Gaseous solvent, gaseous solute
2. Gaseous solvent, solid solute
3. Liquid solvent, gaseous solute
4. Solid solvent, gaseous solute
5. Solid solvent, liquid solute

Answer

b. Gaseous solvent, solid solute

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Exercise \(\PageIndex{1c}\)

Like dissolves like” refers to the fact that _____.

1. Condensed phases can only dissolve other condensed phases
2. Gases can only dissolve other gases
3. Polar solvents dissolve nonpolar solutes and vice versa
4. Polar solvents dissolve polar solutes and nonpolar solvents dissolve nonpolar solutes
5. Solvents can only dissolve solutes of similar molar mass

Answer

d. Polar solvents dissolve polar solutes and nonpolar solvents dissolve nonpolar solutes

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Exercise \(\PageIndex{1d}\)

Which of the following should be immiscible with carbon tetrachloride, CCl₄?

1. Br₂
2. C₃H₈
3. C₆H₁₄
4. CH₃CH₂OH
5. CHCl₃

Answer

d. CH₃CH₂OH,  note:  you might expect chloroform (CHCl₃) to be immiscible because it is polar, but it is miscible in CCl₄.  This is easiest understood by looking at the similarity of the two molecules (like dissolves in like) in that the only difference between them is one of the chlorine atoms is replaced by a hydrogen, which does make it polar, but the large chlorines are highly polarizable and that effect takes precedence.  On an exam we would not give you a borderline case like this, but chloroform is a common solvent and soluble in carbon tetrachloride. 

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Exercise \(\PageIndex{1e}\)

Of the following, _____ will be the most soluble in CCl₄.

1. C₁₀H₂₂
2. CH₃CH₂OH
3. H₂O
4. NaCl
5. NH₃

Answer

a. C₁₀H₂₂

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Exercise \(\PageIndex{1f}\)

_____ is more likely to dissolve in CH₃OH.

1. CCl₄
2. CH₃CH₂OH
3. H₂
4. Kr
5. N₂

Answer

b. CH₃CH₂OH

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Exercise \(\PageIndex{1h}\)

Of the following, _____ is more likely to dissolve in benzene (C₆H₆)?

1. CCl₄
2. CH₃CH₃OH
3. HBr
4. NaCl
5. NH₃

Answer

a. CCl₄

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Exercise \(\PageIndex{1i}\)

Which of the following alcohols is the most soluble in water?

1. CH₃OH
2. CH₃CH₂OH
3. CH₃CH₂CH₂OH
4. CH₃CH₂CH₂CH₂OH
5. CH₃CH₂CH₂CH₂CH₂OH

Answer

a. CH₃OH

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Exercise \(\PageIndex{1j}\)

Which of the following alcohols is the most soluble in hexane (C₆H₁₄)?

1. CH₃OH
2. CH₃CH₂OH
3. CH₃CH₂CH₂OH
4. CH₃CH₂CH₂CH₂OH
5. CH₃CH₂CH₂CH₂CH₂OH

 

Answer

e. CH₃CH₂CH₂CH₂CH₂O

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Exercise \(\PageIndex{1k}\)

A saturated solution _____.

1. cannot be attained
2. contains as much solvent as it can hold
3. contains dissolved solute in equilibrium with undissolved solid
4. contains no double bonds
5. will rapidly precipitate if a seed crystal is added

Answer

c. contains dissolved solute in equilibrium with undissolved solid

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Exercise \(\PageIndex{1l}\)

A supersaturated solution _____.

1. cannot be attained
2. contains as much solvent as it can hold
3. contains dissolved solute in equilibrium with undissolved solid
4. contains no double bonds
5. will rapidly form a precipitate if a seed crystal is added

Answer

e. will rapidly form a precipitate if a seed crystal is added

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Exercise \(\PageIndex{1m}\)

Of the following, which is correctly arranged in order of increasing solubility in water? (Least soluble to most soluble.)

1. LiF < NaNO₃ < CHCl₃
2. CH₄ < NaNO₃ < CHCl₃
3. CH₃OH < CH₄ < LiF
4. CH₃OH < CCl₄ < CHCl₃
5. CCl₄ < CHCl₃ < NaNO₃

Answer

e. CCl₄ < CHCl₃ < NaNO₃

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Exercise \(\PageIndex{1n}\)

In lecture we discussed the aqueous reaction of ferric iron with iodide to form ferrous iron and iodine,

  Fe⁺³  +  I^-  à  Fe⁺²  +  I₂. 

If one were to place the solution in a separatory funnel and add carbon tetrachloride CCl₄, you would expect which of the following to happen?

a.       The solvents would be miscible, with the iodine going into the water layer and the iodide going into the carbon tetrachloride layer.
b.      The solvents would be immiscible, with the iodine going into the water layer and the iodide going into the carbon tetrachloride layer.\
c.       The solvents would be miscible, with the iodide going to the water layer and the iodine going to  the carbon tetrachloride layer.
d.      The solvents would be immiscible, with the iodide going to the water layer and the iodine going to the carbon tetrachloride layer
e.       They would be immiscible forming a homogenous solution with the iodine forming a gas.

Answer

d.      The solvents would be immiscible, with the iodide going to the water layer and the iodine going to the carbon tetrachloride layer

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Units of Concentration

Review of Molarity

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Exercise \(\PageIndex{2a}\)

Calculate the molarity of a solution containing 15.2g H₃AsO₄ in 600mL of water. The molecular weight of H₃AsO₄ is 141.94g/mol.

Answer

\[15.2g H_{3}AsO_{4}*\left ( \frac{1 mol}{141.94g H_{3}AsO_{4}} \right )*\left ( \frac{1}{600mL} \right )*\left ( \frac{1000mL}{1L} \right )=0.178M\]

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Exercise \(\PageIndex{2b}\)

Given the molecular weight of H₃AsO₄ is 141.94g/mol, calculate the mass of H₃AsO₄ required to prepare a 0.75M solution in 400mL of water.

Answer

\[400mL*\left ( \frac{1L}{1000mL} \right )*\left ( \frac{0.75mol}{1L} \right )*\left ( \frac{141.94gH_{3}AsO_{4}}{1mol} \right )=42.6gH_{3}AsO_{4}\]

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Exercise \(\PageIndex{2c}\)

What volume of water must be added to 47 grams of KBr to make a 2.6M solution? The molecular weight of KBr is 119g/mol.

Answer

\[47gKBr*\left ( \frac{1molKBr}{119gKBr} \right )*\left ( \frac{1L}{2.6molKBr} \right )*\left ( \frac{1000mL}{1L} \right )=151.9mL = 150 mL\]

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Exercise \(\PageIndex{2d}\)

How many grams of KBr are required to make 50mL of a 1.2M solution in water? The molecular weight of KBr is 119g/mol.

Answer

\[50mL*\left ( \frac{1L}{1000mL} \right )*\left ( \frac{1.2molKBr}{1L} \right )*\left ( \frac{119gKBr}{1molKBr} \right )=7.14gKBr\]

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Exercise \(\PageIndex{2e}\)

Calculate the molarity of a solution containing 13.41g of AgCl₂ in 250mL of water. The molecular weight of AgCl₂ is 178.8g/mol.

Answer

\[13.41gAgCl_{2}*\left ( \frac{1molAgCl_{2}}{178.8gAgCl_{2}} \right )*\left ( \frac{1}{250mL} \right )*\left ( \frac{1000mL}{1L} \right )=0.300M\]

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Exercise \(\PageIndex{2f}\)

What is the molarity of a Cl^- in a solution prepared by dissolving 23.7g of CaCl₂ in 375g of water and a density of 1.05g/mL?

Answer

\[\left ( \frac{23.7g\,CaCl_{2}}{1} \right )*\left ( \frac{1mol}{110.98g\,CaCl_{2}} \right )*\left ( \frac{1}{375mL} \right )*\left ( \frac{1000mL}{1L} \right )=0.5695M\,CaCl_{2}\]

For each CaCl₂ dissolved two Cl^- ions created so,

\[0.5695M*2=1.14M\,Cl^{-}\]

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Molality 

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Exercise \(\PageIndex{2g}\)

What is the definition of molality?

Answer

moles of solute/kg of solvent

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Exercise \(\PageIndex{2h}\)

Calculate the molality of a solution containing 112g CH₃OH in 134mL of water. The density of water is 1.0g/mL and the molecular weight of CH₃OH is 32.0g/mol.

Answer

\[112gCH_{3}OH*\left ( \frac{1molCH_{3}OH}{32.0gCH_{3}OH} \right )*\left ( \frac{1}{134mL} \right )*\left ( \frac{1mL}{1g} \right )*\left ( \frac{1000g}{1kg} \right )=26.1m\]

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Exercise \(\PageIndex{2i}\)

Calculate the molality of all the ions by adding 1.4g NaCl to 75mL of water.

Answer

\[1.4gNaCl*\left ( \frac{1mol_{NaCl}}{58.44gNaCl} \right )*\left ( \frac{2mol_{ion}}{1mol_{NaCl}} \right )*\left ( \frac{1}{75mL} \right )*\left ( \frac{1mL}{1g} \right )*\left ( \frac{1000g}{1kg} \right )=0.64m\]

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Exercise \(\PageIndex{2j}\)

Calculate the molality of the Br^- ion by adding 2.2g MgBr₂ to 100mL of water. The molecular weight of MgBr₂ is 184.1g/mol.

Answer

\[2.2gMgBr_{2}*\left ( \frac{1molMgBr_{2}}{184.1gMgBr_{2}} \right )*\left ( \frac{2molBr^{-}}{1molMgBr_{2}} \right )*\left ( \frac{1}{100mL} \right )*\left ( \frac{1mL}{1g} \right )*\left ( \frac{1000g}{1kg} \right )=0.24m\]

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Exercise \(\PageIndex{2k}\)

What volume of water in mL would be required to prepare a 32.6m solution from 78g of CaCl₂?

Answer

\[78gCaCl_{2}*\left ( \frac{1molCaCl_{2}}{110.98gCaCl_{2}} \right )*\left ( \frac{1kg}{32.6molCaCl_{2}} \right )*\left ( \frac{1000g}{1kg} \right )*\left ( \frac{1mL}{1g} \right )=21.6mL\]

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Exercise \(\PageIndex{2l}\)

What mass of KBr should be added to 120mL of water to create a 1.8m solution? The molecular weight of KBr is 119g/mol.

Answer

a 1.8m solution of KBr contains  214.2Kg KBr per liter 

(\(1.8 \frac{mol \;KBr}{kg \; H_20}\))(\(\frac{119g \; KBr}{mol}\))=214.8 gKBr/kgH₂O

Water has a density of 1g/mL, so 120mL water = 120g = 0.12Kg,

So  \(  \left ( \frac{214.8g \; KBr}{kg \; H_2O} \right )0.12kg=25.7g \; KBr \)

 

 

Exercise \(\PageIndex{2an}\)

Calculate the molality of a solution containing 56.0g CH₃OH in 134mL of water. The density of water is 1.0g/mL and the molecular weight of CH₃OH is 32.0g/mol.

Answer

13.1m

\[
\text{Molality }(m)=\frac{\text{moles of solute}}{\text{kg of solvent}}
\]

\[
n_{\mathrm{CH_3OH}}=\frac{56.0\ \text{g}}{32.0\ \text{g mol}^{-1}}=1.75\ \text{mol}
\]

Convert the volume of water to mass using density:
\[
m_{\mathrm{H_2O}}=134\ \text{mL}\left(1.0\ \frac{\text{g}}{\text{mL}}\right)=134\ \text{g}
\]

Convert grams of water to kilograms:
\[
m_{\mathrm{H_2O}}=134\ \text{g}\left(\frac{1\ \text{kg}}{1000\ \text{g}}\right)=0.134\ \text{kg}
\]

Now calculate molality:
\[
m=\frac{1.75\ \text{mol}}{0.134\ \text{kg}}=13.1\ \text{mol kg}^{-1}
\]

\[
\boxed{m \approx 13.1\ \text{m}}
\]
 

 

Mole and Mass Fractions

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Exercise \(\PageIndex{2m}\)

Calculate the mole fraction of methane in a 10.0L flask containing 0.367 moles of methane, 0.221 moles hydrogen, and 0.782 moles carbon dioxide.

Answer

\begin{equation}
\text {Mole fraction }=\mathrm{X}_{\mathrm{A}}=\frac{\text { moles } \mathrm{A}}{\text { total number of moles }}
\end{equation}

\begin{equation}
\text { Mole fraction }=\mathrm{X}_{\text {methane}}=\frac{\text { moles methane }}{\text { moles methane }+\text { moles hydrogen }+\text { moles carbon dioxide }}
\end{equation}

\begin{equation}
\text { Mole fraction }=\mathrm{X}_{\text {methane}}=\frac{\text { 0.367 moles methane }}{\text { 0.367 moles methane }+\text { 0.221 moles hydrogen }+\text { 0.782 moles carbon dioxide }}
\end{equation}

\begin{equation}
\text { Mole fraction }=\mathrm{X}_{\text {methane}}=\frac{0.367}{1.37}=0.268
\end{equation}

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Exercise \(\PageIndex{2n}\)

If the mole fraction for nitrogen in a gaseous mixture were 0.565 and the total number of moles present were 2.49 moles, how many moles of nitrogen are present?

Answer

\begin{equation}
\text {Mole fraction }=\mathrm{X}_{\mathrm{A}}=\frac{\text { moles } \mathrm{A}}{\text { total number of moles }}
\end{equation}

\begin{equation}
\text {Mole fraction }=\mathrm{X}_{\text {nitogen }}=\frac{\text { moles nitrogen }}{\text { total number of moles }}
\end{equation}

\begin{equation}
\text {Mole fraction }=0.565=\frac{\text { moles nitrogen }}{\text { 2.49 total moles }}
\end{equation}

\begin{equation}
\text {moles nitrogen }=\text {0.565}*\text {2.49 total moles}
\end{equation}

\begin{equation}
\text {moles nitrogen }=\text {1.41 moles nitrogen}
\end{equation}

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Exercise \(\PageIndex{2o}\)

Calculate the Mass percent if 7.50g CaCl₂ were placed in 500mL of water. The density of water is 1.0g/mL.

Answer

\begin{equation}
\text {Mass fraction}=\frac{\text {mass solute}}{\text {total mass}}
\end{equation}

\begin{equation}
\text { Mass Fraction }=\frac{\text { mass } \mathrm{CaCl}_{2}}{\text { mass } \mathrm{CaCl}_{2}+\text { mass water }}
\end{equation}

\begin{equation}
\text { Mass Fraction }=\frac{7.50 \mathrm{g} \mathrm{CaCl}_{2}}{7.50 \mathrm{g} \mathrm{CaCl}_{2}+(1.0 \mathrm{g} / \mathrm{mL} \text { water } * 500 \mathrm{mL})}
\end{equation}

\begin{equation}
\text { Mass Fraction }=\frac{7.50 \mathrm{g} \mathrm{CaCl}_{2}}{7.50 \mathrm{g} \mathrm{CaCl}_{2}+500 \mathrm{g} \text { water }}
\end{equation}

\begin{equation}
\text { Mass Fraction }=\frac{7.50 \mathrm{g} \mathrm{CaCl}_{2}}{507.5 \mathrm{g}}
\end{equation}

\begin{equation}
\text { Mass Percent }=0.0148 * 100 \%
\end{equation}

\begin{equation}
\text { Mass Percent }=1.48\%
\end{equation}

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Exercise \(\PageIndex{2p}\)

If a scientist needed a NaCl solution that had a mass percent of 2.83%, how many grams of NaCl must be added to 1.00 L of water?

Answer

\[\text{ assume 1 liter water = 1,000g and if }fraction_{NaCl}=0.283 \;the\; fraction_{water}=1-.0283=0.9717\\fraction_{water}=\frac{m_{water}}{m_{water}+m_{NaCl}} \\ m_{NaCl}=\frac{m_{water}(1-fraction_{water})}{fraction_{water}}=\frac{1000g(0.0283)}{0.9717}=29.1g\]

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Exercise \(\PageIndex{2q}\)

If 0.589g of KCl is added to a solvent, a mass fraction NaCl of 0.482 is measured. How many grams of solvent must be present?

Answer

\[f_{NaCl}\equiv fraction_{NaCl} \;\; and \;\;m_{total} \equiv m_T \\ \;\\f_{NaCl}\left ( m_T \right )=m_{NaCl} \therefore\;m_T=\frac{m_{NaCl}}{f_{NaCl}}=\frac{0.589g}{0.482}=1.222g\\ \; \\M_T = m_{NaCl}+m_{solvent} \; \therefore \;\;m_{solvent}=m_T-m_{NaCl}=1.222g-0.589g = 0.633g\]

 

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Exercise \(\PageIndex{2r}\)

What is the mole fraction of NH₃ in a solution prepared by dissolving 15.0g NH₃ in 250g of water with a resulting density of 0.974g/mL?

Answer

\[15.0g\,NH_{3}*\left ( \frac{1mol}{17.031g\,NH_{3}} \right )=0.8807mol\,NH_{3}\]

\[250g\,H_{2}O*\left ( \frac{1mol}{18.015g\,H_{2}O} \right )=13.8773mol\,H_{2}O\]

\[X_{NH_{3}}=\frac{moles\,of\,NH_{3}}{total\,moles}=\frac{0.8807mol\,NH_{3}}{(0.8807mol\,NH_{3}+13.8773mol\,H_{2}O)}=0.0597\]

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Exercise \(\PageIndex{2s}\)

What is the mole fraction of He in gaseous solution containing 4.0g of He, 6.5g of Ar, and 10.0g of Ne?

Answer

\[4.0g\,He*\frac{1mol\,He}{4.0026g\,He}=0.9994mol\,He\]

\[6.5g\,Ar*\frac{1mol\,Ar}{39.948g\,Ar}=0.1627mol\,Ar\]

\[10.0g\,Ne*\frac{1mol\,Ne}{20.180g\,Ne}=0.4955mol\,Ne\]

\[X_{He}=\frac{moles\,of\,He}{total\,moles}=\frac{0.9994mol\,He}{(0.9994mol\,He+0.1627mol\,Ar+0.4955mol\,Ne)}=0.60\]

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Exercise \(\PageIndex{2am}\)

Concentrated aqueous ammonia is 28% by mass NH₃ and has a density of 0.90g/ml. What is the mole fraction ammonia in concentrated ammonia? 

a)  0.58            b) 0.29             c) 0.42             d) 0.81q           e) None of the above

Answer

     b) 0.29  

Given:  
28% by mass NH₃  
ρ = 0.90 g mL⁻¹

Goal:  
Determine the mole fraction of ammonia, \( x_{\mathrm{NH_3}} \).

Strategy:  
Choose a convenient basis.  
Because the composition is given as mass percent, take a basis of 100.0 g of solution.  
(The density is not required for mole fraction.)
 

\[
m_{\mathrm{NH_3}} = 0.28(100.0\ \text{g}) = 28.0\ \text{g}
\]

\[
m_{\mathrm{H_2O}} = 100.0\ \text{g} - 28.0\ \text{g} = 72.0\ \text{g}
\]

Convert masses to moles using molar masses:

\[
M_{\mathrm{NH_3}} = 17.031\ \text{g mol}^{-1},
\qquad
M_{\mathrm{H_2O}} = 18.015\ \text{g mol}^{-1}
\]

\[
n_{\mathrm{NH_3}} =
\frac{28.0\ \text{g}}{17.031\ \text{g mol}^{-1}} = 1.644\ \text{mol}
\]

\[
n_{\mathrm{H_2O}} =
\frac{72.0\ \text{g}}{18.015\ \text{g mol}^{-1}} = 3.997\ \text{mol}
\]

Now compute the mole fraction of ammonia:

\[
x_{\mathrm{NH_3}} =
\frac{1.644}{1.644 + 3.997} = 0.291
\]

\[
\boxed{x_{\mathrm{NH_3}} \approx 0.29}
\]
 

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Parts-per notation

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Exercise \(\PageIndex{2t}\)

Calculate the parts per million of 2.00L of 2.76 * 10^-3 M KMnO₄.  The molecular weight of KMnO₄ is 158g/mol.

Answer

\[Part\,per\,million = \frac{mg\,solute}{L\,solution}\]

\[Part\,per\,million = \frac{mg\,KMnO_{4}}{L\,solution}\]

we are converting from one description of concentration to another, and so the quantity does not matter

\frac{2.76*10^{-3}\,mol}{1\,L\,KMnO_{4}}\left ( \frac{158\,g}{1\,mol\,KMnO_{4}} \right )\left ( \frac{1000\,mg}{1\,g} \right )=\frac{436\,mg\,KMnO_{4}}{L} = 436\,ppm

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Exercise \(\PageIndex{2u}\)

Calculate the parts per million of 2.5L of 7.42 * 10^-3M KCl.  The molecular weight of KCl is 74.55 g/mol.

Answer

\[Part\,per\,million = \frac{mg\,solute}{L\,solution}\]

\[Part\,per\,million = \frac{mg\,KCl}{L\,solution}\]

\[\left ( \frac{7.42*10^{-3}\,mol}{1\,L\,KCl} \right )\left ( \frac{74.55\,g}{1\,mol\,KCl} \right )\left ( \frac{1000\,mg}{1\,g} \right )=553 \frac{mg\,KCl}{L\,solution} = 553ppm\]

 

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Exercise \(\PageIndex{2v}\)

Calculate the parts per billion of 6.2L of a 2.77 * 10^-6M solution of ZnCl₂.  The molecular weight of ZnCl₂ is 136.3g/mol.

Answer

\[Part\,per\,billion = \frac{\mu g\,solute}{L\,solution}\]

\[\mu g\,solute=\mu g\,ZnCl_{2}\]

\[\left ( \frac{2.77*10^{-6}\,mol \,ZnCl_2}{1\,L} \right )*\left ( \frac{136.3\,g \; ZnCl_2}{1\,mol} \right )*\left ( \frac{1*10^{6}\,\mu g}{1\,g} \right )\]

\[\mu g\,ZnCl_{2}=2340.8\,\mu g\,ZnCl_{2}\]

\[Parts\,per\,billion=\frac{2340.8\,\mu g\,ZnCl_{2}}{6.2\,L\,solution}\]

\[Parts\,per\,billion=377\,ppb\]

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Exercise \(\PageIndex{2w}\)

Calculate the parts per billion of 5.0L of a 2.762 * 10^-8M solution of AgBr₂.

Answer

\[Part\,per\,billion = \frac{\mu g\,solute}{L\,solution}\]

\[\mu g\,solute=\mu g\,AgBr_{2}\]

\[\mu g\,AgBr_{2}=5.0\,L*\left ( \frac{2.762*10^{-8}\,mol}{1\,L} \right )\left ( \frac{267.7\,g}{1\,mol} \right )\left ( \frac{1*10^{6}\,\mu g}{1\,g} \right )\]

\[\mu g\,AgBr_{2}=36.97\, \mu g\,AgBr_{2}\]

\[Parts\,per\,billion=\frac{36.97\,\mu g\,AgBr_{2}}{5.0\,L\,solution}\]

\[Parts\,per\,billion=7.39\,ppb\]

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Mass Percent and Molarity

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Exercise \(\PageIndex{2x}\)

Which of the following is correct about a solution containing 28% phosphoric acid by mass?

1. The density of this solution is 2.8g/mL
2. 100g of this solution contains 28g of phosphoric acid
3. 1 mL of this solution contains 28g of phosphoric acid
4. 1 L of this solution has a mass of 28g
5. 1 L of this solution contains 28mL of phosphoric acid

Answer

b. 100g of this solution contains 28g of phosphoric acid

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Exercise \(\PageIndex{2y}\)

What is the concentration of chloride ion (mass %) in a solution that contains 35.0ppm chloride?

Answer

\[35ppm=\frac{35mg}{L} \approx \frac{35mg}{10^6mg}=3.5x10^{-5}=fraction_{chloride} \text{ (as the mass of the solute is neglible compared to water, with density = 1g/mL)} \\ \therefore \\ \; \\ \% \;chloride=100f_{chloride}=0.0035\% \text{ chloride by mass} \nonumber\]

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Exercise \(\PageIndex{2z}\)

What is the concentration of CaCl₂ (mass %), if a solution is prepared by dissolving 23.7g of CaCl₂ in 375g of water with a density of 1.05g/mL?

Answer

\[\%\,by\,mass=\frac{mass\,of\,solute}{total\,mass\,of\,solution}*100\]

\[\%\,by\,mass=\frac{23.7g}{23.7g\,+\,375g}*100=5.94\%\]

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Exercise \(\PageIndex{2aa}\)

Urea has a molar mass of 60.0g/mol. If 16g of urea is dissolved in 39g of water, what is the resulting concentration in mass percent?

Answer

\[\%\,by\,mass=\frac{mass\,of\,solute}{total\,mass\,of\,solution}*100\]

\[\%\,by\,mass=\frac{16g}{16g\,+\,39g}*100=29\%\]

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Exercise \(\PageIndex{2ab}\)

An organic solution is 72% water by mass and has a density of 1.3 g/mL.  What is the molarity of the solution if the molecular weight of the solute is 64 g/mol. 

Answer

Molarity = \(\frac{mole_{solute}}{L_{solution}}\)

If you have 1 liter the total mass is 1,300g as there are 1,000 mL in a L

If it is 72% water it is 28% solute, so the mass fraction solute is 0.28, and the mass solute in 1 liter is 0.28(1,300)=364g

So the moles solute in 1 liter is 364g(\(\frac{mole solute}{64g}\))=5.6875mol

So the molarity is 5.7mol/L=5.7M

 

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Exercise \(\PageIndex{2ac}\)

Commercial vinegar has been analytically determined to be 10.2% acetic acid (CH₃COOH) by mass.  The molecular mass of acetic acid is 60.0 g/mol.  The density of vinegar is 1.01g/mL.  Calculate the molarity of acetic acid in 50.0 mL of vinegar.

Answer

\[Molarity=\frac{moles\,CH_{3}COOH}{L\,solution}\]

\[Molarity=\left ( \frac{10.2g\,CH_{3}COOH}{100g\,solution} \right )*\left ( \frac{1.01g\,solution}{1mL\,solution} \right )*\left ( \frac{1mol\,CH_{3}COOH}{60g\,CH_{3}COOH} \right )*\left ( \frac{1000mL}{L} \right )\]

\[Molarity=1.72M\]

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Exercise \(\PageIndex{2ad}\)

What is the molarity of a solution that is 3.60% by mass NaOH.  The molecular mass of NaOH is 40.0 g/mol, and the density of the solution is 1.00g/mL.

Answer

\[Molarity=\frac{moles\,NaOH}{L\,solution}\]

\[Molarity=\left ( \frac{3.6g\,NaOH}{100g\,solution} \right )*\left ( \frac{1.00g\,solution}{1mL\,solution} \right )*\left ( \frac{1mol\,NaOH}{40g\,NaOH} \right )*\left ( \frac{1000mL}{L} \right )\]

\[Molarity=0.900M\]

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Exercise \(\PageIndex{2ae}\)

A 2.7M solution is found to be composed of 8.3% by mass of an unknown organic compound.  The density of the solution was experimentally determined to be 1.5g/mL.  In order for a scientist to identify the compound, he must know the molecular mass.  Calculate the molecular mass of the unknown compound.

Answer

\[fw_{uk}=\frac{m(g)}{n(mole)}_{uk} \text{and from the molarity we know one liter has 2.7 moles, starting with fraction solute and keeping units in expression} \\ \; \\ \frac{8.3g_{solute}}{100g_{solution}}\left ( \frac{1,500g}{L} \right )_{solution}\left ( \frac{L_{solution}}{2.7mol_{solute}} \right )=46\frac{g}{mol}_{solute}\]

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Exercise \(\PageIndex{2af}\)

Calculate the molar concentration of a solution containing 11.8g MgBr₂ dissolved in 100mL of water.  The density of the solution is 1.2g/mL, and the molecular mass of MgBr₂ is 184g/mol.

Answer

\[Find \; M_{MgBr_2}=\frac{n_{MgBr_2}}{L_{solution}} \text{start with the mass percent (mass solute into total mass) of your sample and convert} \; \\ \; \frac{11.8g_{MgBr_2}}{111.8g_{solution}} \left ( \frac{1200g_{solution}}{L{solution}} \right )\left ( \frac{1 mol_{MgBr_2}}{184g} \right )=0.69M\]

 

General Questions

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Exercise \(\PageIndex{2ag}\)

A solution of 26.5g of methanol (CH₃OH) is dissolved in 244g of water with a density of 0.975g/mL. What is the molarity, molality, and mole fraction of methanol in the solution?

Molarity

\[Find \; M_{CH_3OH}=\frac{n_{CH_3OH}}{L_{solution}} \text{ start with the mass percent} \; \\ \; \; \\  \frac{26.5g_{CH_3OH}}{\left (26.5g_{CH_3OH}+244g_{H_2O} \right )} \left ( \frac{975g_{solution}}{L{solution}} \right )\left ( \frac{1 mol_{CH_3OH}}{32.04g} \right )=2.98M\]

Molality

\[Find \; m_{CH_3OH}=\frac{n_{CH_3OH}}{Kg_{H_2O}} \text{ start with the mass ratio} \; \\ \; \frac{26.5g_{CH_3OH}}{0.244Kg_{H_2O}} \left ( \frac{1 mol}{32.04g} \right )_{CH_3OH}=3.39M\]

Mole fraction

\[\text{start with mass fraction and convert to mole fraction} \\ \; \\ \; \frac{26.5g\left ( \frac{mol}{32.04g} \right )}{26.5g\left ( \frac{mol}{32.04g} \right )+244g\left ( \frac{mol}{18.0154g} \right )}=0.0576\]

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Exercise \(\PageIndex{2ah}\)

Calculate the molarity of an aqueous sodium chloride solution that is 13.0% sodium chloride by mass and density of 1.10g/mL.

Answer

\[Find \; M_{NaCl}=\frac{n_{NaCl}}{L_{solution}} \text{ start with the mass fraction} \; \\ \; \\ \frac{13.0g_{NaCl}}{100g_{solution}}\left ( \frac{1100g_{solution}}{L{solution}} \right )\left ( \frac{1 mol}{58.44g} \right )_{NaCl}=2.45M\]

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Exercise \(\PageIndex{3.2ai}\)

Calculate the percent composition by mass of sodium chloride in an aqueous solution of sodium chloride with a concentration of 2.23M and density of 1.01g/mL.

Answer

\[\text{Consider 1 liter, calculate the mass fraction and multiply by 100} \; \\ \; \\ \left ( \frac{2.23mole_{NaCl}}{L_{solution}}\left ( \frac{58.44g}{mol} \right )_{NaCl}\left ( \frac{1L}{1,010g} \right )_{solution} \right )100=12.9 \%\]

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Exercise \(\PageIndex{2aj}\)

An aqueous solution of sodium chloride has a concentration of 2.22M and 11.0% sodium chloride by mass. What is the density (g/mL) of this solution?

Answer

 

\[d=\frac{m_{total}}{V} ,\;\;M=\frac{n_{solute}}{V} \\consider \; 1 \;liter \; n_{solute}=M(V)=\frac{2.2mol}{V}(1L)=2.2 mol\\fw=\frac{m_{solute}}{n_{solute}} \therefore \;m_{solute}=n{solute}(fw_{solute})=2.2mol(58.44\frac{g}{mol})=128.568g\\m_{solute}=fraction_{solute}m_{total}\;\;\therefore \;\;m_{total}=\frac{m_{solute}}{fract_{solute}}=\frac{128.568}{0.110}=1168.8\\d=\frac{1170g}{L}=1.17g/mL\]

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Exercise \(\PageIndex{2ak}\)

We learned from the "solubility rules" (sec. 3.4.3) in gen chem 1 that silver chloride is an insoluble salt here.  A saturated solution of silver chloride at 50^oC has a concentration of 520 \(\mu\)g AgCl/100 g H₂O.  What is the largest possible mole fraction of silver chloride?

Answer

\[X_{AgCl}=\frac{520x10^{-6}gAgCl\left ( \frac{mol\, AgCl}{143.32g} \right )}{\left ( 520x10^{-6}gAgCl\left ( \frac{mol\, AgCl}{143.32g} \right ) \right )+100gH_2O\left (\frac{ mol\, H_2O}{18.015g} \right )}=0.00000065\]    .

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Exercise \(\PageIndex{2al}\)

We learned from the "solubility rules" (sec. 3.4.3) in gen chem 1 that sodium chloride is an soluble salt here.  A saturated solution of silver chloride at 25^oC has a concentration of 36g NaCl/100 g H₂O.  What is the largest possible mole fraction of sodium chloride?

Answer

\[X_{AgCl}=\frac{36.0gNaCl\left ( \frac{mol\, NaCl}{58.443g} \right )}{\left ( 36gNaCl\left ( \frac{mol\, NaCl}{58.443^{\circ}g} \right ) \right )+100gH_2O\left (\frac{ mol\, H_2O}{18.015g} \right )}=0.0999\]    .

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Solvation Process

Exercise \(\PageIndex{3A}\)

_____ is the process of solute particles being surrounded by solvent particles.

1. Agglomeration
2. Agglutination
3. Passivation
4. Salutation
5. Solvation

Answer

e. Solvation

 

 

Pressure and Temperature Effects on Solubility

 

Exercise \(\PageIndex{4a}\)

The solubility of a gas in a liquid increases when….

a.  the partial gas pressure is reduced and the temperature is lowered.

b.  the partial gas pressure is reduced and the temperature is raised.

c.  the partial gas pressure is increased and the temperature is lowered.

d.  the partial gas pressure is increased and the temperature is raised

e.  the partial gas pressure is increased but it is unaffected by changes in temperature

Answer

c.  the partial gas pressure is increased and the temperature is lowered.

 

Henry's Law

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Exercise \(\PageIndex{4b}\)

The solubility of nitrogen in blood is 5.4x10^-4M at 37°C and 0.8atm.  When a deep-sea diver breathes the compressed air from a tank at a partial pressure of 1.6atm, what would be the solubility of nitrogen if the temperature remains constant?

Answer

\[S=kP\]

\[\frac{S_{1}}{S_{2}}=\frac{kP_{1}}{kP_{2}}\]

\[S_{1}=S_{2}\frac{P_{1}}{P_{2}}\]

\[S_{1}=5.4*10^{-4}M\left ( \frac{1.6atm}{0.8atm} \right )\]

\[S_{1}=1.08*10^{-3}M\]

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Exercise \(\PageIndex{4c}\)

What is the Henry’s law Constant k in the above question?

Answer

\[S=kP\]

\[\frac{S}{P}=k=\frac{5.4*10^{-4}M}{0.8atm}=6.75*10^{-4}M/atm\]

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Exercise \(\PageIndex{4d}\)

The solubility of CO₂ in water is 0.035M at 25°C and 1 atm.  What is the Henry’s law Constant k?

Answer

\[S=kP\]

\[\frac{S}{P}=k=\frac{0.035M}{1atm}=0.035M/atm\]

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Exercise \(\PageIndex{4e}\)

What would be the solubility of CO₂ in water at 0.04atm?  (Use the information from Q 13.3.4)

Answer

\[S=kP\]

\[\frac{S_{1}}{S_{2}}=\frac{kP_{1}}{kP_{2}}\]

\[S_{1}=S_{2}\frac{P_{1}}{P_{2}}\]

\[S_{1}=0.035M\left ( \frac{0.04atm}{1atm} \right )\]

\[S_{1}=1.4*10^{-3}M\]

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Exercise \(\PageIndex{4f}\)

The solubility of oxygen in water at 0°C and 1 atm is 0.07g/L, what is the Henry’s law Constant k?

Answer

\[S=kP\]

\[\frac{S}{P}=k=\frac{0.07g/L}{1atm}=0.07g/L*atm\]

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Exercise \(\PageIndex{4g}\)

What would be the solubility of oxygen in water at 2.5atm? (Use the information from Q 13.3.6)

Answer

\[S=kP\]

\[\frac{S_{1}}{S_{2}}=\frac{kP_{1}}{kP_{2}}\]

\[S_{1}=S_{2}\frac{P_{1}}{P_{2}}\]

\[S_{1}=0.07g/L\left ( \frac{2.5atm}{1atm} \right )\]

\[S_{1}=0.175g/L\]

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Exercise \(\PageIndex{4h}\)

What is the concentration of nitrogen in water when the partial pressure of N₂ above the solution is 0.826 atm? Henry’s Law constant for this system is 6.8*10^-4 mol/L*atm.

Answer

\[S=kP\]

\[S=kP=\left (6.8*10^{-4}mol/L*atm \right )*\left ( 0.826atm \right )=5.6*10^{-4}M\]

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Exercise \(\PageIndex{4i}\)

The solubility of oxygen gas in water at 25°C and 1.0 atm pressure of oxygen is 0.041 g/L. The solubility of oxygen in water at 3.0 atm and 25°C is _____ g/L.

Answer

\[S=kP\]

\[\frac{S_{1}}{S_{2}}=\frac{kP_{1}}{kP_{2}}\]

\[\frac{S_{1}}{S_{2}}=\frac{P_{1}}{P_{2}}\]

\[S_{2}=S_{1}*\frac{P_{2}}{P_{1}}=0.041g/L*\frac{3.0atm}{1.0atm}=0.12g/L\]

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Exercise \(\PageIndex{4j}\)

The solubility of argon in water is 25°C is 1.6*10^-3 mol/L when the pressure of argon above the solution is 1.0 atm. The solubility of argon at a pressure of 2.5 atm is _____ mol/L.

Answer

\[S=kP\]

\[\frac{S_{1}}{S_{2}}=\frac{kP_{1}}{kP_{2}}\]

\[\frac{S_{1}}{S_{2}}=\frac{P_{1}}{P_{2}}\]

\[S_{2}=S_{1}*\frac{P_{2}}{P_{1}}=\left ( 1.6*10^{-3}mol/L \right )*\frac{2.5atm}{1.0atm}=4.0*10^{-3}mol/L\]

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Exercise \(\PageIndex{4k}\)

 

The solubility of nitrogen gas in water at 25°C and 0.78 atm is 5.3*10^-4 M. What is the partial pressure of nitrogen when its solubility in water (at 25°C) is 1.1*10^-3 M?

Answer

\[S=kP\]

\[\frac{S_{1}}{S_{2}}=\frac{kP_{1}}{kP_{2}}\]

\[\frac{S_{1}}{S_{2}}=\frac{P_{1}}{P_{2}}\]

\[P_{2}=P_{1}*\frac{S_{2}}{S_{1}}=0.78atm*\frac{1.1*10^{-3}M}{5.3*10^{-4}M}=1.6atm\]

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Exercise \(\PageIndex{4l}\)

The solubility of_____ when dissolved in liquids are affected by their partial pressure.

1. Gases
2. Solids
3. Liquids
4. All of the above
5. Solids and liquids

Answer

a. Gases

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Exercise \(\PageIndex{2ao}\)

The solubility of argon in water is 25°C is 1.6*10^-3 mol/L when the pressure of argon above the solution is 1.0 atm. The solubility of argon at a pressure of 5.0 atm is _____ mol/L.

Answer

\[
S = k_H P
\]

\[
\frac{S_2}{S_1} = \frac{P_2}{P_1}
\]

\[
\frac{S_2}{1.6\times10^{-3}\ \text{mol L}^{-1}}
=
\frac{5.0\ \text{atm}}{1.0\ \text{atm}}
\]

\[
S_2 =
\left(1.6\times10^{-3}\ \text{mol L}^{-1}\right)
\left(\frac{5.0}{1.0}\right)
\]

\[
S_2 = 8.0\times10^{-3}\ \text{mol L}^{-1}
\]

\[
\boxed{S_2 = 8.0\times10^{-3}\ \text{mol L}^{-1}}
\]
 

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13.4: Colligative Properties

Boiling Point

 

Exercise \(\PageIndex{5a}\)

The boiling point elevation constant for water is 0.52 C/m. A 1.34 m aqueous solution of compound X had a boiling point of 101.4C. Which of the following could be compound X?

1. Na₃PO₄
2. KCl
3. CH₃CH₂OH
4. CaCl₂
5. C₆H₁₂O₆

Answer

b. KCl

 

Exercise \(\PageIndex{5b}\)

Ethylene glycol is a common component of anti-freeze.  A solution of ethylene glycol and water has a boiling point of 101.0°C.  What is the molality of the solution?  K_bp of water is 0.5121°C /m

Answer

\[\Delta t=K_{bp}m\]

The boiling point of pure water is 100°C.

\[m=\frac{\Delta t}{K_{bp}}\]

\[m=\frac{101.0\,^{0}C-100.00\,^{0}C}{0.5121\,^{0}C/m}\]

\[m=1.95m\]

 

Exercise \(\PageIndex{5c}\)

How many grams of ethylene glycol (MW=62.068 g/mol) is need to add to 250g of water to increase the boiling point by 1.00°C? K_bp of water is 0.5121°C /m

Answer

\[\Delta t=K_{bp}m\]

\[m=\frac{\Delta t}{K_{bp}}\]

\[m=\frac{1.0\,^{0}C}{0.5121\,^{0}C/m}\]

\[m=1.95m\]

\[molality=\frac{moles\,of\,the\,solute}{mass\,of\,the\,solvent\,in\,kg}\]

Assume there is x grams fo ethylene glycol,

\[\frac{\frac{x}{62.068}}{\frac{250}{1000}}=1.95m\]

solve for x, x=30.3g

 

Exercise \(\PageIndex{5d}\)

Determine the boiling point of a 0.800m solution of CaF₂. 

Answer

K_bp of water is 0.5121°C/m

The solution of CaF₂ is a mixture of water and CaF₂, a strong electrolyte. As each CaF₂ creates one Ca⁺² and two F^-, so for an ideal solution the Van't Hoff factor is 3 (the total molality is 2.40m).

\[\Delta t=iK_{bp}m=3*0.5121*0.800=1.23\,^{0}C\]

\[T=100.0\,^{0}C+1.23\,^{0}C=101.23\,^{0}C\]

 

Exercise \(\PageIndex{5e}\)

Methyl Salicylate is used in food flavoring and pharmaceuticals. Adding 1.25g of methyl salicylate (MW=152.14g/mol) to 100g of benzene will raise the boiling point to what temperature?  K_bp of benzene is 2.53°C /m, and the boiling point of pure benzene is 80.10°C.

Answer

\[\Delta t=K_{bp}m\]

\[molality=\frac{moles\,of\,the\,solute}{mass\,of\,the\,solvent\,in\,kg}\]

\[\Delta t=\frac{\frac{1.25g}{152.14g/mol}}{\frac{100g}{1000}}*2.53\,^{0}C/m=0.21\,^{0}C\]

\[T=80.10\,^{0}C+0.21\,^{0}C=80.31\,^{0}C\]

 

Exercise \(\PageIndex{5f}\)

Adding 36.0g of glucose (MW= 180g/mol) to 500.0 g of ethyl alcohol will raise the boiling point to 79.3°C.  Pure ethanol has a boiling point of 78.5°C.  What is the value for K_bp?

Answer

\[\Delta t=K_{bp}m\]

\[K_{bp}=\frac{\Delta t}{m}=\frac{79.3-78.5}{\frac{\frac{36.0}{180}}{0.500}}=2.0\,^{0}C/m\]

 

Freezing Point Depression

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Exercise \(\PageIndex{5g}\)

Which of the following liquids have the lowest freezing point?

1. Pure H₂O
2. Aq. 0.050 m glucose
3. Aq. 0.030 m NaI
4. Aq. 0.030 m CoI₂
5. Aq. 0.030 m AlI₃

Answer

e. Aq. 0.030 m AlI₃

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Exercise \(\PageIndex{5h}\)

What is the freezing point of an aqueous 2.00m NaCl solution? K_f of water is 1.86°C/m.

Answer

\[\Delta t=K_{f}m\]

The solution of Nacl is a mixture of water and NaCl, a strong electrolyte. As each NaCl creates one Na+ and one Cl- and if we ignore ion pairing (Van't Hoff effect) the so 2.00m NaCl s 4.00m in all ions.

\[\Delta t=K_{f}m=1.86*4.00=7.44\,^{0}C]

The freezing point of pure water is 0°C

\[t=0-7.44=-7.44\,^{0}C\]

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Exercise \(\PageIndex{5i}\)

Determine the freezing point of a solution that is made of 3.60g of glucose C₆H₁₂O₆ in 20.0g of water. K_f of water is 1.86°C/m.

Answer

\[\Delta t=K_{f}m\]

\[m=\frac{\frac{3.60g}{180g/mol}}{0.0200kg}=1.00m\]

The freezing point of pure water is 0°C

\[t=0-1.86=-1.86\,^{0}C\]

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Exercise \(\PageIndex{5j}\)

Reserpine is a substance that is used as a tranquilizer and sedative. A solution containing 0.20g of reserpine in 5.0g of camphor (K_f = 40.0°C/m) gives a freezing point 2.63°C below the freezing point of pure camphor. Determine the molecular weight of reserpine.

Answer

\[\Delta t=K_{f}m\]

\[m=\frac{\Delta t}{K_{f}}=\frac{2.63}{40.0}=0.06575\]

\[m=\frac{\frac{0.20g}{Molar\,mass}}{0.050kg}=0.06575\]

\[Molar\,mass=608.4g/mol\]

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Exercise \(\PageIndex{5k}\)

The melting point of pure benzene is 278.70K and K_f 4.90K/m. When 2.10g of an unknown solute is added to 50.0g of benzene, the freezing point of the solution is 277.60K. Determine the molecular weight of the unknown.

Answer

\[\Delta t=278.70-277.60=1.10\]

\[m=\frac{\Delta t}{K_{f}}=\frac{1.10}{4.90}=0.224\]

\[m=\frac{\frac{2.10g}{Molar\,mass}}{0.0500kg}=0.224\]

\[Molar\,mass=187.09g/mol\]

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Exercise \(\PageIndex{5l}\)

The freezing point of a solution prepared with 3.54g of acetamide (C₂H₅ON) to 200.0g of naphthalene (C₁₀H₈) was 78.2°C.  Pure naphthalene solidified at 80.2°C.  Determine the freezing point constant for naphthalene.

Answer

\[\text{Note, m is for molality and mass, so we will use g for the mass solute and kg for the mass solvent and m for molality}\]

\[\Delta T = -K_{fp}m \;\;\;  fw=\frac{g_{solute}}{n_{solute}} \;\;\; m_{solution}=\frac{g_{solute}}{kg_{solvent}} \]

\[\Delta T=78.2-80.2=-2.0\]

\[m=\frac{\frac{3.54g}{59g/mol}}{0.200kg}=0.30\]

\[K_{f}=\frac{\Delta t}{m}=\frac{2.0}{0.30}=6.7K/m\]

 

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Exercise \(\PageIndex{5m}\)

How many grams of antifreeze (C₂H₆O₂) need be added to 200.0g of water to have a solution that will not freeze at the temperature -10°C?  K_f of water is 1.86°C/m. How many grams of antifreeze (C₂H₆O₂) need be added to 200.0g of water to have a solution that will not freeze at the temperature -10°C?  K_f of water is 1.86°C/m

Answer

\[\Delta t=0-(-10.0)=10.0\]

\[m=\frac{\Delta t}{}K_{f}=\frac{10.0}{1.86}=5.38\]

\[m=\frac{\frac{mass}{62g/mol}}{0.200kg}=5.38\]

\[mass=66.7g\]

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Exercise \(\PageIndex{5n}\)

Calculate the freezing point (°C) of a solution prepared by dissolving 50.0g of glycerin (C₃H₈O₃, a nonelectrolyte) in 200g of ethanol. Ethanol has a freezing point of -114.6°C and a molal freezing point depression constant of 2.00 °C/m.

Answer

\[moles\,solute=50.0g*\frac{1mol}{92.064g}=0.543mol\]

\[kg\,solvent=200g*\frac{1kg}{1000g}=0.200kg\]

\[m=\frac{moles\,solute}{kg\,solvent}=\frac{0.543mol}{0.200kg}=2.72m\]

\[\Delta t=K_{f}m=2.00\,^{0}C/m*2.72m=5.43\,^{0}C\]

\[\Delta t=T_{f\,solute}-T_{f\,solution}\]

\[T_{f\,solution}=T_{f\,solute}-\Delta t=-114.6\,^{0}C-5.42\,^{0}C=-120.0\,^{0}C\]

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Exercise \(\PageIndex{5ag}\)

A solution made by dissolving 3.54 g of an unknown solute in 200.0 g of naphthalene that freezes at 78.2 °C, while pure naphthalene freezes at 80.2 °C.
Given K_fp(naphthalene) = 6.67 °C·kg·mol^{-1}, find the formula weight (FW) of the solute

Answer

 

\[
\Delta T = T_f^\circ(\text{pure}) - T_f(\text{solution}) = 80.2 - 78.2 = 2.0
\]

\[
\Delta T = -K_{fp}\,m
\qquad
fw = \frac{g_{solute}}{n_{solute}}
\qquad
m = \frac{n_{solute}}{kg_{solvent}}
\]

 Substitute n_solute into fw using molality ---
\[
n_{solute} = m\,kg_{solvent}
\quad\Rightarrow\quad
fw = \frac{g_{solute}}{m\,kg_{solvent}}
\]

Solve the FP-depression equation for m, then substitute ---
\[
\Delta T = -K_{fp}\,m
\quad\Rightarrow\quad
m = \frac{-\Delta T}{K_{fp}}
\]

\[
\Rightarrow\quad
fw = \frac{g_{solute}}{\left(\dfrac{-\Delta T}{K_{fp}}\right)\,kg_{solvent}}
= -\,\frac{g_{solute}\,K_{fp}}{kg_{solvent}\,\Delta T}
\]

Plug numbers: g_solute = 3.54 g, kg_solvent = 0.2000 kg, ΔT = -2.0 or use |ΔT|=2.0, Kfp=6.67 ---
\[
fw
= -\,\frac{(3.54)\,(6.67)}{(0.2000)\,(-2.0)}
\ \mathrm{g\,mol^{-1}}
= \frac{3.54\times 6.67}{0.4000}
\ \mathrm{g\,mol^{-1}}
\approx 59.0\ \mathrm{g\,mol^{-1}}.
\]

\[
\boxed{fw(\text{solute}) \approx 5.90\times 10^{1}\ \mathrm{g\,mol^{-1}}}
\]
 

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Exercise \(\PageIndex{5ai}\)

A solution made by dissolving 3.54 g of an unknown solute in 200.0 g of naphthalene that freezes at 78.2 °C, while pure naphthalene freezes at 80.2 °C.
Given K_fp(naphthalene) = 6.67 °C·kg·mol^-1, find the formula weight (FW) of the solute in units of g/mol.

Answer

59.0 g/mol

\[
\Delta T_f = T_f^\circ - T_f
\]

\[
\Delta T_f = 80.2^\circ\text{C} - 78.2^\circ\text{C} = 2.0^\circ\text{C}
\]

\[
\Delta T_f = K_f\,m
\]

\[
m = \frac{\Delta T_f}{K_f} = \frac{2.0^\circ\text{C}}{6.67^\circ\text{C}\cdot \text{kg mol}^{-1}} = 0.300\ \text{mol kg}^{-1}
\]

\[
m = \frac{n_{\text{solute}}}{\text{kg solvent}}
\]

\[
n_{\text{solute}} = m(\text{kg solvent}) = (0.300\ \text{mol kg}^{-1})(0.2000\ \text{kg}) = 0.0600\ \text{mol}
\]

\[
\text{FW} = \frac{m_{\text{solute}}}{n_{\text{solute}}} =
\frac{3.54\ \text{g}}{0.0600\ \text{mol}} = 59.0\ \text{g mol}^{-1}
\]

\[
\boxed{\text{FW} \approx 59.0\ \text{g mol}^{-1}}
\]
 

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Vapor Pressure and Raoult's Law 

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Exercise \(\PageIndex{5o}\)

Which of the following solutes will have the lowest vapor pressure in a 0.100m solution?

1. NaCl
2. CCl₄
3. CH₃OH
4. Ca(ClO₄)₂
5. Al(ClO₄)₃

Answer

e. Al(ClO₄)₃

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Exercise \(\PageIndex{5p}\)

When _____ negative deviations from Raoult’s Law are encountered.

1. Solute-solute and solvent-solvent interactions are stronger than solute-solvent interactions,
2. Interactions between solute and solvent are very weak,
3. Solute-solute and solvent-solvent interactions have the same strength as solute-solvent interactions,
4. None of these
5. Interactions between solute and solvent are exceptionally strong,

Answer

b. Interactions between solute and solvent are very weak,

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Exercise \(\PageIndex{5q}\)

The vapor pressure of water at room temperature is 23.8 torr.  If 90g of glucose (C₆H₁₂O₆) were added to 500.0 g of water, what will be the vapor pressure of the solution?    

Answer

\[P_{solvent}=X_{solvent}*P^{0}_{solvent}\]

\[90g\,C_{6}H_{12}O_{6}*\left ( \frac{1mol\,C_{6}H_{12}O_{6}}{180g\,C_{6}H_{12}O_{6}} \right )=0.5mol\,C_{6}H_{12}O_{6}\]

\[500g\,H_{2}O*\left ( \frac{1mol\,H_{2}O}{18g\,H_{2}O} \right )=27.8mol\,\,H_{2}O\]

\[X=\frac{27.8mol\,H_{2}O}{27.8mol\,H_{2}O+0.5mol\,C_{6}H_{12}O_{6}}=0.98\]

\[P=0.98*23.8=23.3torr\]

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Exercise \(\PageIndex{5r}\)

What would be the vapor pressure when 32.0g of naphthalene(C₁₀H₈) are dissolved in 1000g of benzene (C₆H₆) at 25⁰C?  The vapor pressure of benzene at 25⁰C is 110 torr.

Answer

\[P_{solvent}=X_{solvent}*P^{0}_{solvent}\]

The molar mass of naphthalene is 128g/mol, and he benzene is 78g/mol.

\[P=\frac{\frac{1000}{78}}{\frac{1000}{78}+\frac{32}{128}}*110torr=107.9torr\]

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Exercise \(\PageIndex{5s}\)

The vapor pressure of water at 30.0⁰C is 31.8 torr.  How many moles of glucose C₆H₁₂O₆  need to be added into 150.0 g of water to lower the vapor pressure to 29.0 torr?

Answer

\[P_{solvent}=X_{solvent}*P^{0}_{solvent}\]

\[X_{solvent}=\frac{P_{solvent}}{P^{0}_{solvent}}=\frac{29.0}{31.8}=0.91\]

\[\frac{150.0g}{18g/mol}=8.3moles\,of\,water\]

\[X_{solvent}=\frac{8.3}{8.3+moles\,of\,glucose}=0.91\]

\[moles\,of\,glucose:\,0.8moles\]

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Exercise \(\PageIndex{5t}\)

When 90.0 g of the solid were added to 50.0g of water, the vapor pressure of the solution dropped from 17.4 torr to 12.4 torr.  Determine the molecular mass of the compound. 

Method 1

\[P_{solvent}=X_{solvent}*P^{0}_{solvent}\]

\[X_{solvent}=\frac{P_{solvent}}{P^{0}_{solvent}}=\frac{12.4}{17.4}=\frac{\frac{50.0}{18}}{\frac{50.0}{18}+\frac{90.0}{MW}}\]

solve for MW,

\[MW=80.4g/mol\]

Method 2

\[\Delta P=-X_{solute}*P^{0}_{solvent}\]

\[\Delta P=12.4-17.4=-5.0torr\]

\[X_{solvent}=-\frac{\Delta P}{P^{0}_{solvent}}=-\frac{-5.0}{17.4}=0.287\]

\[X_{solvent}=\frac{\frac{90.0}{MW}}{\frac{90.0}{MW}+\frac{50.0}{18}}=0.287\]

solve for MW,

\[MW=80.4g/mol\]

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Exercise \(\PageIndex{5u}\)

What is the vapor pressure of a 10% aqueous solution of sucrose (C₁₂H₂₂O₁₁) at 20⁰C?  The vapor pressure of water at 20⁰C is 17.5 torr.

Answer

Assume the solution is 100.0g. Then, 10% of the solution implies that the solution contains 10.0g of the solute (sucrose MW=342g/mol), and 90.0g of water.

\[\frac{10.0g}{342g/mol}=0.029mol\,sucrose\]

\[\frac{90.0}{18g/mol}=5.0g\]

\[X=\frac{5.0}{5.0+0.029}=0.994\]

\[P_{solvent}=X_{solvent}*P^{0}_{solvent}\]

\[P=0.994*17.5=17.4torr\]

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Exercise \(\PageIndex{5v}\)

Pure ethanol has a vapor pressure of 349 torr at 60°C. Using Raoult’s Law predict the vapor pressure of 10.0 mmol naphthalene (nonvolatile) dissolving in 90.0 mmol of ethanol.

Answer

\[X=\frac{moles\,ethanol}{total\,moles}=\frac{90.0mmol}{90.0mmol+10.0mmol}=0.90\]

\[P=X*P^{0}=0.90*349torr=314torr\]

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Exercise \(\PageIndex{5w}\)

Calculate the vapor pressure of water above a solution prepared by dissolving 18g of glucose (a nonelectrolyte, MW=180g/mol) in 95g of water? Pure water has a vapor pressure of 23.8 torr at 25°C.

Answer

\[18g\,glucose*\frac{1mol}{180g\,glucose}=0.1mol\,glucose\]

\[95g\,water*\frac{1mol}{18g\,water}=5.28mol\,water\]

\[X=\frac{moles\,water}{total\,moles}=\frac{5.28mol}{5.28mol+0.1mol}=0.981\]

\[P=X*P^{0}=0.981*23.8torr=23.4torr\]

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Exercise \(\PageIndex{5ah}\)

 What would be the vapor pressure when 32.0g of naphthalene(C₁₀H₈) are dissolved in 1000g of benzene (C₆H₆) at 25⁰C?  The vapor pressure of benzene at 25⁰C is 110 torr.

Answer

108 torr     

\[
P_{\text{solution}} = X_{\mathrm{benzene}}\,P^\circ_{\mathrm{benzene}}
\]

\[
X_{\mathrm{benzene}}=\frac{n_{\mathrm{benzene}}}{n_{\mathrm{benzene}}+n_{\mathrm{naphthalene}}}
\]

\[
M_{\mathrm{C_6H_6}} = 6(12.01)+6(1.008)=78.11\ \text{g mol}^{-1}
\qquad
M_{\mathrm{C_{10}H_8}} = 10(12.01)+8(1.008)=128.17\ \text{g mol}^{-1}
\]

\[
n_{\mathrm{benzene}}=\frac{1000\ \text{g}}{78.11\ \text{g mol}^{-1}}=12.80\ \text{mol}
\]

\[
n_{\mathrm{naphthalene}}=\frac{32.0\ \text{g}}{128.17\ \text{g mol}^{-1}}=0.250\ \text{mol}
\]

\[
X_{\mathrm{benzene}}=
\frac{12.80}{12.80+0.250}=0.9809
\]

\[
P_{\text{solution}}=(0.9809)(110\ \text{torr})=1.079\times 10^{2}\ \text{torr}
\]

\[
\boxed{P_{\text{solution}} \approx 108\ \text{torr}}
\]
 

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Osmotic Pressure

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Exercise \(\PageIndex{5x}\)

What is the osmotic pressure of a solution made by adding 9.0g of glucose, C₆H₁₂O₆, to enough water to make 250ml solution at 25°C?

Answer

\(\pi =MRT\), where R = 0.08206 L/K*mol

\[M=\frac{\frac{9.0g}{180g/mol}}{0.250L}=0.2M\]

\[\pi =0.2*0.08206*(273.15+25)=4.89atm\]

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ercise \(\PageIndex{5y}\)

The osmotic pressure of blood is 7.5atm at 37°C.  How much glucose should be used per liter to make an intravenous injection that has the same osmotic pressure as blood?

Answer

\(\pi =MRT\), where R = 0.08206 L/K*mol

\[M=\frac{\pi }{RT}=\frac{7.5}{0.08206*310.15}=0.295M\]

\[0.95mol/L*180g/mol=53.04g\]

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Exercise \(\PageIndex{5z}\)

What is the osmotic pressure of 0.0015g of NaCl at 25°C?

Answer

\(\pi =MRT\), where R = 0.08206 L/K*mol

The solution of NaCl is a mixture of water and NaCl, a strong electrolyte. As each NaCl creates on Na+ and one Cl- and if we ignore ion pairing (Van't Hoff effect) the 0.0015M NaCl is 0.0030M in all ions.

\[\pi =0.0030*0.08206*298.15=0.074atm\]

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Exercise \(\PageIndex{5aa}\)

What is the minimum pressure required to desalinate 1.0M NaCl solution at 25°C?

Answer

\(\pi =MRT\), where R = 0.08206 L/K*mol

The solution of NaCl is a mixture of water and NaCl, a strong electrolyte. As each NaCl creates on Na+ and one Cl- and if we ignore ion pairing (Van't Hoff effect) the 1.0M NaCl is 2.0M in all ions.

\[\pi =2.0*0.08206*298.15=49atm\]

The reverse osmotic pressure has to be twice as much to desalinate the salt solution. Therefore, the answer will be 98 atm.

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Exercise \(\PageIndex{5ab}\)

What is the osmotic pressure of a 0.25M urea solution that is isotonic with seawater at 4°C?

Answer

\(\pi =MRT\), where R = 0.08206 L/K*mol

\[\pi =0.25*0.08206*277.15=5.7atm\]

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Exercise \(\PageIndex{5ac}\)

The osmotic pressure of a tryptophan solution is 103.5torr at 25°C.  If the solution has a density of 1.136g/L, what is the molar mass of the amino acid?

Answer

\(\pi =MRT\), where R = 0.08206 L/K*mol

\[103.5torr*\frac{1atm}{760torr}=0.136atm\]

\[M=\frac{\pi }{RT}=\frac{0.136}{0.08206*298.15}=0.0056M\]

\[\frac{\frac{1.136g}{molar\,mass}}{1L}=0.0056M\]

\[molar\,mass=204g/mol\]

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Exercise \(\PageIndex{5ad}\)

Using a solution prepared by dissolving 0.60g of nicotine (a nonelectrolyte) in water to make 12mL of solution has an osmotic pressure of 7.55atm at 25C. What is the molecular weight of nicotine in g/mol?

Answer

\(\pi =MRT\), where R = 0.08206 L/K*mol

\[M=\frac{\pi }{RT}=\frac{7.55atm}{0.08206*(25+273.15)}=0.309M\]

\[12mL*\frac{1L}{1000mL}\frac{0.309mol}{1L}=3.70*10^{-3}mol\]

\[3.70*10^{-3}mol=\frac{0.60g}{molar\,mass}\]

\[molar\,mass=162g/mol\]

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Mixed Questions

Exercise \(\PageIndex{ae}\)

The identity of the _____ determines the magnitude of the k_fp and k_bp.

1. Temperature
2. Solvent
3. Solution
4. Solute and solvent
5. Solute

Answer

b. Solvent

 

Exercise \(\PageIndex{af}\)

The _____ decreases when adding solute to the solution.

1. Vapor pressure
2. Osmotic pressure
3. Freezing point and vapor pressure
4. Freezing point
5. Boiling point

Answer

c. Freezing point and vapor pressure

 

Colloidial Mixtures

 

General Questions

 

This part is under construction

 

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Exercise \(\PageIndex{18}\)

What is the process of a substance sticking to the surface of another?

1. Effusion
2. Diffusion
3. Coagulation
4. Adsorption
5. Absorption

Answer

d. Adsorption

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Exercise \(\PageIndex{19}\)

Which of the following produces the greatest number of ion when one mole dissolves in water?

1. Sucrose
2. NH₄NO₃
3. NH₄Cl
4. NaCl
5. Na₂SO₄

Answer

e. Na₂SO₄

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Exercise \(\PageIndex{20}\)

Consider an aqueous solution of a nonvolatile compound, the vapor pressure will be _____, the boiling point will be _____, and the freezing point will be _____ than for pure water.

1. Lower, lower, lower
2. Lower, higher, lower
3. Lower, higher, higher
4. Higher, lower, higher
5. Higher, higher, lower

Answer

b. Lower, higher, lower

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