PROBLEM \(\PageIndex{1}\)
What mass of a concentrated solution of nitric acid (68.0% HNO3 by mass) is needed to prepare 400.0 g of a 10.0% solution of HNO3 by mass?
- Answer
-
58.8 g
PROBLEM \(\PageIndex{2}\)
What mass of a 4.00% NaOH solution by mass contains 15.0 g of NaOH?
- Answer
-
375 g
- Click here to see a video of the solution
-
PROBLEM \(\PageIndex{3}\)
What mass of solid NaOH (97.0% NaOH by mass) is required to prepare 1.00 L of a 10.0% solution of NaOH by mass? The density of the 10.0% solution is 1.109 g/mL.
- Answer
-
\(\mathrm{114 \;g}\)
PROBLEM \(\PageIndex{4}\)
The hardness of water (hardness count) is usually expressed in parts per million (by mass) of \(\ce{CaCO_3}\), which is equivalent to milligrams of \(\ce{CaCO_3}\) per liter of water. What is the molar concentration of Ca2+ ions in a water sample with a hardness count of 175 mg CaCO3/L?
- Answer
-
\(1.75 \times 10^{−3} M\)
PROBLEM \(\PageIndex{5}\)
A throat spray is 1.40% by mass phenol, \(\ce{C_6H_5OH}\), in water. If the solution has a density of 0.9956 g/mL, calculate the molarity of the solution.
- Answer
-
0.148 M
- Click here to see a video of the solution
-
PROBLEM \(\PageIndex{6}\)
Copper(I) iodide (CuI) is often added to table salt as a dietary source of iodine. How many moles of CuI are contained in 1.00 lb (454 g) of table salt containing 0.0100% CuI by mass?
- Answer
-
\(\mathrm{2.38 \times 10^{−4}\: mol}\)
PROBLEM \(\PageIndex{7}\)
What are the mole fractions of H3PO4 and water in a solution of 14.5 g of H3PO4 in 125 g of water?
- Answer
-
\(X_\mathrm{H_3PO_4}=0.021\)
\(X_\mathrm{H_2O}=0.979\)
- Click here to see a video of the solution
-
PROBLEM \(\PageIndex{8}\)
What are the mole fractions of HNO3 and water in a concentrated solution of nitric acid (68.0% HNO3 by mass)?
- Answer
-
\(X_\mathrm{HNO_3}=0.378\)
\(X_\mathrm{H_2O}=0.622\)
PROBLEM \(\PageIndex{9}\)
Calculate the mole fraction of each solute and solvent:
- 583 g of H2SO4 in 1.50 kg of water—the acid solution used in an automobile battery
- 0.86 g of NaCl in 1.00 × 102 g of water—a solution of sodium chloride for intravenous injection
- 46.85 g of codeine, C18H21NO3, in 125.5 g of ethanol, C2H5OH
- 25 g of I2 in 125 g of ethanol, C2H5OH
- Answer a
-
\(X_\mathrm{H_2SO_4}=0.067\)
\(X_\mathrm{H_2O}=0.933\)
- Answer b
-
\(X_\mathrm{HCl}=0.0026\)
\(X_\mathrm{H_2O}=0.9974\)
- Answer c
-
\(X_\mathrm{codiene}=0.054\)
\(X_\mathrm{EtOH}=0.946\)
- Answer d
-
\(X_\mathrm{I_2}=0.035\)
\(X_\mathrm{EtOH}=0.965\)
- Click here to see a video of the solution
-
PROBLEM \(\PageIndex{10}\)
Calculate the mole fraction of each solute and solvent:
- 0.710 kg of sodium carbonate (washing soda), Na2CO3, in 10.0 kg of water—a saturated solution at 0 °C
- 125 g of NH4NO3 in 275 g of water—a mixture used to make an instant ice pack
- 25 g of Cl2 in 125 g of dichloromethane, CH2Cl2
- 0.372 g of histamine, C5H9N, in 125 g of chloroform, CHCl3
- Answer a
-
\(X_\mathrm{Na_2CO_3}=0.0119\)
\(X_\mathrm{H_2O}=0.988\)
- Answer b
-
\(X_\mathrm{NH_4NO_3}=0.09927\)
\(X_\mathrm{H_2O}=0.907\)
- Answer c
-
\(X_\mathrm{Cl_2}=0.192\)
\(X_\mathrm{CH_2CI_2}=0.808\)
- Answer d
-
\(X_\mathrm{C_5H_9N}=0.00426\)
\(X_\mathrm{CHCl_3}=0.997\)
PROBLEM \(\PageIndex{11}\)
What is the difference between a 1 M solution and a 1 m solution?
- Answer
-
In a 1 M solution, the mole is contained in exactly 1 L of solution. In a 1 m solution, the mole is contained in exactly 1 kg of solvent.
PROBLEM \(\PageIndex{12}\)
What is the molality of phosphoric acid, H3PO4, in a solution of 14.5 g of H3PO4 in 125 g of water?
- Answer
-
1.18 m
PROBLEM \(\PageIndex{13}\)
What is the molality of nitric acid in a concentrated solution of nitric acid (68.0% HNO3 by mass)?
- Answer
-
33.7 m
PROBLEM \(\PageIndex{14}\)
Calculate the molality of each of the following solutions:
- 0.710 kg of sodium carbonate (washing soda), Na2CO3, in 10.0 kg of water—a saturated solution at 0°C
- 125 g of NH4NO3 in 275 g of water—a mixture used to make an instant ice pack
- 25 g of Cl2 in 125 g of dichloromethane, CH2Cl2
- 0.372 g of histamine, C5H9N, in 125 g of chloroform, CHCl3
- Answer
-
6.70 × 10−1 m
- Answer
-
5.67 m
- Answer
-
2.8 m
- Answer
-
0.0358 m
PROBLEM \(\PageIndex{15}\)
A 13.0% solution of K2CO3 by mass has a density of 1.09 g/cm3. Calculate the molality of the solution.
- Answer
-
1.08 m
Feedback
Think one of the answers above is wrong? Let us know here.