2.14 Practice Problems
Attempt these problems as if they were real exam questions in an exam environment.
Only look up information if you get severely stuck. Never look at the solution until you have exhausted all efforts to solve the problem. Having to look up information or the solution should be an indicator that the previous layers (1–5) in the Structured Learning Approach have not been mastered.
At 20 °C, the concentration of dissolved O2 in a sample of water at a partial pressure of 101.3 kPa is 1.38 × 10–3 mol L–1. What is the solubility of O2 (in mol L–1) when the partial pressure is 234.5 kPa?
- 1.36 × 10–5
- 2.82 × 10–4
- 5.49 × 10–2
- 3.19 × 10–3
- 6.022 × 1023
Solution
Answer: D
Concept: Henry’s Law
\[C_{\mathrm{g}} = k_{\mathrm{H}} P_{\mathrm{g}}\]
We are asked to find the solubility (here, C) of O2 in water at 234.5 kPa given the solubilty of O2 in water at 101.3 kPa. We need to determine Henry’s constant for O2 using the given information and then use that constant to find the solubility at 234.5 kPa.
Find Henry’s Constant for O2
\[\begin{align*} C_{\mathrm{g}} &= k_{\mathrm{H}} P_{\mathrm{g}} \longrightarrow \\[1.5ex] k_{\mathrm{H}} &= \dfrac{C_{\mathrm{g}}}{P_{\mathrm{g}}} \\[1.5ex] &= \dfrac{1.38\times 10^{-3}~\mathrm{mol~L^{-1}}}{101.3~\mathrm{kPa}} \\[1.5ex] &= 1.3623\times 10^{-5}~\mathrm{mol~L^{-1}~kPa^{-1}} \end{align*}\]
Find Solubility of O2 at 234.5 kPa
\[\begin{align*} C_{\mathrm{g}} &= k_{\mathrm{H}} P_{\mathrm{g}} \\[1.5ex] &= \left ( 1.3623\times 10^{-5}~\mathrm{mol~L^{-1}~kPa^{-1}} \right ) \left ( 234.5~\mathrm{kPa} \right ) \\[1.5ex] &= 3.194\times 10^{-3}~\mathrm{mol~L^{-1}} \end{align*}\]
Which of the following concentration units is temperature dependent?
- χsolute
- m
- mol %
- M
- none of these
Solution
Answer: D
Concept: Concentration Units
Recall that mass is generally temperature independent (mass changes with temperature under normal conditions are trivial). Temperature has a more dramatic effect on volume, however.
A 0.5 L aqueous solution contains 4.32 × 1024 Fe atoms. What is the percent by mass of Fe in this solution? dFe = 7.86 g cm–3
- 22
- 47
- 68
- 74
- 99
Solution
Answer: B
Concept: Mass %
Mass percent is given as
\[\mathrm{mass~\%~solute} = \omega_{\mathrm{solute}} \times 100\%\]
We are starting with the total number of solute atoms in solution. We must convert atoms → moles → mass since ω is mass fraction.
Find Mass of Fe
\[\begin{align*} m &= 4.32\times10^{24}~\mathrm{Fe~atoms} \left ( \dfrac{1~\mathrm{mol}}{6.022\times 10^{23}~\mathrm{Fe~atoms}}\right ) \left ( \dfrac{55.845~\mathrm{g}}{\mathrm{mol}} \right ) \\[1.5ex] &= 400.6151~\mathrm{g~Fe} \end{align*}\]
Find Mass of Water
We know the volume of solution to be 0.5 L. Since we have the density of iron, we can convert mass to volume for iron and subtract the volume of the iron from the solution leaving us with the volume of the water.
\[\begin{align*} V_{\mathrm{solute}} &= \dfrac{m}{d} \\[1.5ex] &= 400.6151~\mathrm{g~Fe} \left ( \dfrac{\mathrm{mL}}{7.86~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \\[2.5ex] &= 0.05097~\mathrm{L}\\[2.5ex] V_{\mathrm{solvent}} &= V_{\mathrm{total}} - V_{\mathrm{solute}} \\[1.5ex] &= 0.5~\mathrm{L} - 0.05097~\mathrm{L} \\[1.5ex] &= 0.44903~\mathrm{L} \end{align*}\]
Since the density of water was not given, we assume a density of 1.0 g mL–1 and find the mass of water.
\[\begin{align*} m_{\mathrm{solvent}} &= dV \\[1.5ex] &= (1.0~\mathrm{g~mL^{-1}})(0.44903~\mathrm{L}) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right )\\[1.5ex] &= 449.03~\mathrm{g} \end{align*}\]
Find Mass of Solution
\[\begin{align*} m_{\mathrm{solution}} &= m_{\mathrm{solute}} + m_{\mathrm{solvent}} \\[1.5ex] &= 400.6151~\mathrm{g~Fe} + 449.03~\mathrm{g~water} \\[1.5ex] &= 849.6451~\mathrm{g} \end{align*}\]
Find Mass Fraction
\[\begin{align*} \omega_{\mathrm{solute}} &= \dfrac{m_{\mathrm{solute}}}{m_{\mathrm{solution}}} \\[1.5ex] &= \dfrac{400.6151~\mathrm{g~Fe}}{849.6451~\mathrm{g~solution}} \\[1.5ex] &= 0.47151 \end{align*}\]
Find Mass %
\[\begin{align*} \mathrm{mass~\%~solute} &= \omega_{\mathrm{solute}} \times 100\% \\[1.5ex] &= 0.47151 \times 100\% \\[1.5ex] &= 47.15\% \end{align*}\]
150.0 g of glucose (C6H12O6; m.m. = 180.16 g mol–1), a nonvolatile solute, is dissolved in 950 g of water at 25 °C. What is the vapor pressure of solution (in atm)? P°H2O = 0.0313 atm.
- 0
- 3.08 × 10–2
- 3.10 × 10–2
- 3.24 × 10–2
- 4.38 × 102
Solution
Answer: B
Concept: Vapor Pressure Lowering
The vapor pressure of a solution of sugar water should be lower than that of pure water and the relation is given by Raoult’s Law.
\[P_{\mathrm{A}} = \chi_{\mathrm{A}} P^{\circ}_{\mathrm{A}}\]
Notice that we’ll need the mole fraction, χ, of the solvent first since it is the only thing that contributes to the vapor pressure of the solution.
Find Moles of Solute
\[\begin{align*} n_{\mathrm{solute}} &= m_{\mathrm{solute}} \times \mathrm{m.m.~solute} \\[1.5ex] &= 150.0~\mathrm{g~glucose} \left ( \dfrac{\mathrm{mol}}{180.16~\mathrm{g}} \right ) \\[1.5ex] &= 0.83259~\mathrm{mol~glucose} \end{align*}\]
Find Moles of Solvent
\[\begin{align*} n_{\mathrm{solvent}} &= m_{\mathrm{solvent}} \times \mathrm{m.m.~solvent} \\[1.5ex] &= 950~\mathrm{g~water} \left ( \dfrac{\mathrm{mol}}{18.02~\mathrm{g}} \right ) \\[1.5ex] &= 52.7192~\mathrm{mol~water} \end{align*}\]
Find Moles of Solution
\[\begin{align*} n_{\mathrm{solution}} &= n_{\mathrm{solute}} + n_{\mathrm{solvent}} \\[1.5ex] &= 0.83259~\mathrm{mol~glucose} + 52.7192~\mathrm{mol~water} \\[1.5ex] &= 53.55179~\mathrm{mol~solution} \end{align*}\]
Find Mole Fraction of Solvent
\[\begin{align*} \chi_{\mathrm{solvent}} &= \dfrac{n_{\mathrm{solvent}}}{n_{\mathrm{solution}}} \\[1.5ex] &= \dfrac{52.7192~\mathrm{mol~water}}{53.55179~\mathrm{mol~solution}} \\[1.5ex] &= 0.9844526 \end{align*}\]
Find Vapor Pressure of Solution
\[\begin{align*} P_{\mathrm{A}} &= \chi_{\mathrm{A}} P^{\circ}_{\mathrm{A}} \\[1.5ex] &= \left ( 0.9844526 \right ) \left ( 0.0313~\mathrm{atm} \right )\\[1.5ex] &= 3.081\times 10^{-2}~\mathrm{atm} \end{align*}\]
- 0
150.0 g of glucose (C6H12O6; m.m. = 180.16 g mol–1), a nonvolatile solute, is dissolved in 950 g of water at 25 °C. What is the vapor pressure of solute (in atm)? P°H2O = 0.0313 atm.
- 0
- 3.08 × 10–2
- 3.10 × 10–2
- 3.24 × 10–2
- 4.38 × 102
Solution
Answer: A
Concept: Vapor Pressure and Volatility
The question is asking for the vapor pressure of the solute (here, glucose). Nonvolatile substances have approximately zero vapor pressure. Therefore, the answer is 0.
20.0 g of benzene (C6H6) and 20.0 g of ethanol (C2H6O), both volatile substances, are added to 100.0 g of water at 25 °C. What is the vapor pressure (in atm) of the solution?
P°water = 0.0313 atm, P°benzene = 0.132 atm, P°ethanol = 0.077 atm- –0.024
- 3.88 × 10–3
- 0.036
- 0.039
- 0.414
Solution
Answer: D
Concept: Vapor Pressure Lowering
The vapor pressure of the solution can be approximated as the sum of the vapor pressure of the volatile solutes and solvent. We need to determine the vapor pressure for each and then add them together such that
\[P_{\text{solution}} = \sum_{i=1}^{n}\left(P_{\text {solute}}\right)_{i}+P_{\text {solvent}}\]
To find the vapor pressure of each component, we use Raoult’s Law.
\[P_{\mathrm{A}} = \chi_{\mathrm{A}} P^{\circ}_{\mathrm{A}}\]
Find Moles of Each Component
\[\begin{align*} n_{\mathrm{water}} &= 100.0~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{18.02~\mathrm{g}} \right ) = 5.5494~\mathrm{mol~water}\\[1.5ex] n_{\mathrm{benzene}} &= 20.0~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{78.11~\mathrm{g}} \right ) = 0.25605~\mathrm{mol~benzene}\\[1.5ex] n_{\mathrm{ethanol}} &= 20.0~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{46.07~\mathrm{g}} \right ) = 0.43412~\mathrm{mol~ethanol} \end{align*}\]
Find Moles of Solution
\[\begin{align*} n_{\mathrm{solution}} &= n_{\mathrm{water}} + n_{\mathrm{benzene}} + n_{\mathrm{ethanol}} \\[1.5ex] &= 5.5494~\mathrm{mol~water} + 0.25605~\mathrm{mol~benzene} + 0.43412~\mathrm{mol~ethanol} \\[1.5ex] &= 6.23957~\mathrm{mol} \end{align*}\]
Find Mole Fraction for Each Component
\[\begin{align*} \chi &= \dfrac{n_i}{n_{\mathrm{tot}}} \\[2ex] \chi_{\mathrm{water}} &= \dfrac{5.5494~\mathrm{mol~H_2O}}{6.23957~\mathrm{mol~solution}} = 0.88939 \\[1.5ex] \chi_{\mathrm{benzene}} &= \dfrac{0.25605~\mathrm{mol~C_6H_6}}{6.23957~\mathrm{mol~solution}} = 0.04104 \\[1.5ex] \chi_{\mathrm{ethanol}} &= \dfrac{0.43412~\mathrm{mol~C_2H_6O}}{6.23957~\mathrm{mol~solution}} = 0.069575 \end{align*}\]
Find Vapor Pressure of Each Component
\[\begin{align*} P &= \chi P^{\circ}\\[2ex] P_{\mathrm{water}} &= \left ( 0.88939 \right ) \left ( 0.0313~\mathrm{atm} \right ) = 0.02784~\mathrm{atm} \\ P_{\mathrm{benzene}} &= \left ( 0.04104 \right ) \left ( 0.132~\mathrm{atm} \right ) = 0.0054173~\mathrm{atm} \\ P_{\mathrm{ethanol}} &= \left ( 0.069575 \right ) \left ( 0.077~\mathrm{atm} \right ) = 0.0053573~\mathrm{atm} \end{align*}\]
Find Vapor Pressure of Solution
\[\begin{align*} P_{\text{solution}} &= \sum_{i=1}^{n}\left(P_{\text{solute}}\right)_{i}+P_{\text{solvent}} \\[1.5ex] &= P_{\mathrm{benzene}} + P_{\mathrm{ethanol}} + P_{\mathrm{water}} \\[1.5ex] &= 0.0054173~\mathrm{atm~benzene} + 0.0053573~\mathrm{atm~ethanol} + 0.02784~\mathrm{atm~water}\\[1.5ex] &= 0.0386146~\mathrm{atm} \end{align*}\]
According to the Starkville 2015 Drinking Water Quality Report, barium was found to exist at a 0.1127 ppm concentration (by mass). If the sample of drinking water that was tested had a mass of 0.25 kg (about 0.250 L), what is the mass (in g) of barium that was found?
- 2.8 × 10–5
- 1.1 × 10–4
- 1.8 × 10–3
- 2.4 × 10–2
- 3.2 × 10–1
Solution
Answer: A
Concept: Concentration Units
We are given a concentration by mass in ppm and need to find the mass of the solute.
Find Mass of Solute
\[\begin{align*} \mathrm{pp}_x &= \dfrac{m_{\mathrm{solute}}}{m_{\mathrm{solution}}} \times \mathrm{~multiplication~factor} \\[2.0ex] \mathrm{ppm} &= \dfrac{m_{\mathrm{solute}}}{m_{\mathrm{solution}}} \times 10^6 \longrightarrow \\[1.5ex] m_{\mathrm{solute}} &= \dfrac{\mathrm{ppm}}{10^6} \times m_{\mathrm{solution}} \\[1.5ex] &= \dfrac{0.1127~\mathrm{ppm}}{10^6} \left ( 0.25~\mathrm{kg~solution} \right ) \left ( \dfrac{10^3~\mathrm{g}}{1~\mathrm{kg}} \right ) \\[1.5ex] &= 2.82\times 10^{-5}~\mathrm{g} \end{align*}\]
Which type of solute in water leads to the most significant change in colligative properties for that solution?
- strong electrolyte
- weak acid
- nonelectrolyte
Solution
Answer: A
Concept: Electrolytes
Strong electrolytes dissociate more than weak electrolytes. Non-electrolytes do not dissociate at all. The solution with the most dissolved particles should experience the greatest change in a colligative property, assuming the sovlent is the same as well as the molal concentration.
Which equation or principle relates the vapor pressure of a solution to its concentration?
- Clausius-Clapeyron
- Henry’s Law
- van’t Hoff Factor
- Raoult’s Law
- none of these
Solution
Answer: D
Concept: Raoult’s Law
Raoult’s Law relates vapor pressure of a solution to its concentration.
\[P_{\mathrm{solution}} = \chi_i P_{\mathrm{solvent}}^{\circ}\]
Which of the following aqueous solutions would have an experimental (i.e. measured) Van’t Hoff factor that is closest to the expected Van’t Hoff factor?
- 0.01 M NaCl
- 0.10 M NaCl
- 0.25 M NaCl
- 0.40 M NaCl
- 1.05 M NaCl
Solution
Answer: A
Concept: van’t Hoff Factor
Dissolve a soluble salt in water. We know that the salt will dissociate to some degree. In dilute solutions, we generally assume all the salt dissociates into its constituent ions (cations and anions) giving us the ideal van’t Hoff factor (iideal).
We know that positive and negative charges attract, so if the oppositely charged solute ions “find each other” in solution, they have may reassociate, thereby lowering the ratio of dissolved particles to the original amount of salt introduced. This is the real/measured van’t Hoff factor (ireal). The “reassociation” process is known as “ion pairing”.
Ions are more likely to “find each other” when there are more ions in solution. Therefore, ireal decreases with increasing concentration. The solution that will have iideal i*real will be the most dilute solution.
What is the expected van’t Hoff factor for a weak, monoprotic acid (HA) that is only 18% dissociated when dissolved in water?
- 0.89
- 1.00
- 1.04
- 1.18
- 2.48
Solution
Answer: D
Concept: van’t Hoff Factor
A weak acid is a weak electrolyte and therefore, a sample of weak acid will not completely dissociate in solution (much of it remains in its associated state).
The problem says only 18% dissociates. Recall that the van’t Hoff factor is a ratio of dissociated particles in solution to the number of particles introduced into solution.
For a non-electrolyte (one that does not dissociate at all), there is 0% dissociation and the the ratio is 1:1 and a van’t Hoff factor of 1.
For a strong, binary electrolyte, there is approximately 100% dissociation leading to a 1:2 ratio or a van’t Hoff factor of 2.
For this weak electrolyte, only 18% dissociates, therefore the ratio is 1:1.18 giving a van’t Hoff factor of 1.18 which is reasonable as it lies between 1 and 2.
Which of the following is a weak electrolyte?
- AgNO3
- C6H6
- Br2
- HF
- none of the above
Solution
Which of the following is a strong electrolyte?
- Fe
- C3H8
- HBr
- H3PO4
- none of these
Solution
250.0 mL of a 1.5 M KCl aqueous solution is dissolved in 500.0 mL of pure water (at 25 °C). What is the final concentration (in M) of KCl?
- 0.05 M
- 0.50 M
- 0.75 M
- 1.10 M
- 4.50 M
Solution
Answer: B
Concept: Concentration Units
This problem demonstrates a simple dilution where a stock KCl solution is diluted with solvent.
Find Volume of Dilute Solution
\[\begin{align*} V_{\mathrm{solution}} &= V_{\mathrm{KCl~solution}} + V_{\mathrm{pure~water}} \\[1.5ex] &= 250.0~\mathrm{mL} + 500.0~\mathrm{mL} \\[1.5ex] &= 750.0~\mathrm{mL} \end{align*}\]
Find Molar Concentration
Use the dilution equation here.
\[\begin{align*} M_1V_1 &= M_2V_2 \\[1.5ex] M_2 &= \dfrac{M_1V_1}{V_2} \\[1.5ex] &= \dfrac{1.5~M \left ( 250.0~\mathrm{mL} \right ) \left ( \dfrac{1~\mathrm{L}}{10^3~\mathrm{mL}} \right )} {750.0~\mathrm{mL} \left ( \dfrac{1~\mathrm{L}}{10^3~\mathrm{mL}} \right )} \\[1.5ex] &= 0.5~M \end{align*}\]
Generally, concentrated hydrofluoric acid (m.m. = 20.01 g mol–1) has a concentration of 48% by mass. What is the molal concentration of HF?
- 10.34
- 4.23
- 23.99
- 2.4 × 10–3
- 46.13
Solution
Answer: E
Concept: Concentration Units
We are given the mass % of the solution. We must convert this to a molal concentration.
Find Mass Fraction
\[\begin{align*} \mathrm{mass~\%} &= \omega_{\mathrm{solute}} \times 100 \longrightarrow \\[1.5ex] \omega_{\mathrm{solute}} &= \dfrac{\mathrm{mass~\%}}{100} \\[1.5ex] &= \dfrac{48}{100} \\[1.5ex] &= 0.48 \end{align*}\]
Find Mass of Solute
Let us assume we have 100 g of solution. Therefore,
\[\begin{align*} \omega_{\mathrm{solute}} &= \dfrac{m_{\mathrm{solute}}}{m_{\mathrm{solution}}} \longrightarrow \\[1.5ex] m_{\mathrm{solute}} &= \omega_{\mathrm{solute}} \times m_{\mathrm{solution}} \\[1.5ex] &= (0.48) (100~\mathrm{g}) \\[1.5ex] &= 48~\mathrm{g} \end{align*}\]
Find Moles of Solute
\[\begin{align*} n_{\mathrm{solute}} &= \dfrac{m}{\mathrm{molar~mass}} \\[1.5ex] &= \dfrac{48~\mathrm{g}}{20.01~\mathrm{g~mol^{-1}}} \\[1.5ex] &= 2.399~\mathrm{mol} \end{align*}\]
Find Molal Concentration
\[\begin{align*} m &= \dfrac{\mathrm{mol~solute}}{\mathrm{kg~solvent}} \\ &= \dfrac{2.399~\mathrm{mol}}{52~\mathrm{g} \left ( \dfrac{\mathrm{kg}}{10^3~\mathrm{g}} \right )} \\[1.5ex] &= 46.13~\mathrm{mol~kg^{-1}} \end{align*}\]
Which of the following aqueous solutions should have the highest boiling point?
- 1.0 m KCl
- 0.50 m NaBr
- 0.50 m K2SO4
- 0.50 m CaBr2
- 1.5 m C6H12O6
Solution
Answer: A
Concept: Boiling Point Elevation
The more particles there are in solution, the higher the boiling point should be as given by
\[\Delta T_{\mathrm{b}} = i m K_{\mathrm{b}}\]
Simply identify the van’t Hoff factor for each solute and then multiply by the given molal concentration (i × m). We assume completely dissociation if any dissociation takes place. The largest value should lead to the largest ΔTb.
KCl: 2 × 1.0 m = 2.0
NaBr: 2 × 0.50 m = 1.0
K2SO4: 3 × 0.50 m = 1.5
CaBr2: 3 × 0.50 m = 1.5
C6H12O6: 1 × 1.5 m = 1.5KCl has the largest value.
A 2.34 L aqueous solution contains 30% by volume of acetic acid (CH3COOH). What is the volume of acetic acid (in mL) in the solution?
- 0.351
- 0.702
- 0.705
- 351
- 702
Solution
Answer: E
Concept: Concentration Units
We must find the volume of the solute in a solution where we are given the concentration as a percent by volume.
\[\dfrac{V_\mathrm{{solute}}}{V_\mathrm{{solution}}} \times \mathrm{~multiplication~factor}\]
Find Volume of Solute
\[\begin{align*} \mathrm{pp}_x &= \dfrac{V_{\mathrm{solute}}}{V_{\mathrm{solution}}} \times \mathrm{~multiplication~factor} \\[2.0ex] \mathrm{vol\%} &= \dfrac{V_{\mathrm{solute}}}{V_{\mathrm{solution}}} \times 100 \longrightarrow \\[1.5ex] V_{\mathrm{solute}} &= \dfrac{\mathrm{vol\%}}{100} \times V_{\mathrm{solution}} \\[1.5ex] &= \dfrac{30}{100} \left ( 2.34~\mathrm{L~solution} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{1~\mathrm{L}} \right ) \\[1.5ex] &= 702~\mathrm{mL} \end{align*}\]
The freezing point of a 0.75 m solution of chloroform (CHCl3) is 3.1 °C lower than that of the pure solvent. What is the freezing point of a similar solution (in °C) containing 500 g of H2O and 0.1667 mol of CHCl3?
- –6.38
- –2.481
- –2.267
- –1.378
- 2.25
Solution
Answer: D
Concept: Freezing Point Depression
We break this problem down into two parts. The first part has us solve for the cryoscopic constant, Kf for a solvent. We will need to use this constant in the second part of the problem since the solvent doesn’t change, only the concentration of the solute. Also, the van’t Hoff factor for chloroform is 1 since it is not a salt or acid/base.
Find Cryoscopic Constant of Solute
\[\begin{align*} \Delta T_{\mathrm{f}} &= i m K_{\mathrm{f}} \longrightarrow \\[1.5ex] K_{\mathrm{f}} &= \dfrac{\Delta T_{\mathrm{f}}}{im} \\[1.5ex] &= \dfrac{(3.1~^{\circ}\mathrm{C}}{1 \times 0.75~m} \\[1.5ex] &= 4.133~^{\circ}\mathrm{C}~m^{-1} \end{align*}\]
Note that the change in freezing point is a positive value numerically (Kf is a positive value) but we know that this corresponds to a decrease in the freezing point.
Find Molal Concentration of New Solution
\[\begin{align*} m &= \dfrac{n_{\mathrm{solute}}}{\mathrm{kg~solvent}} \\[1.5ex] &= \dfrac{0.1667~\mathrm{mol~CHCl_3}}{500~\mathrm{g~H_2O} \left ( \dfrac{1~\mathrm{kg}}{10^3~\mathrm{g}}\right )}\\[1.5ex] &= 0.3334~\mathrm{mol~kg^{-1}} \end{align*}\]
Find Freezing Point of New Solution
\[\begin{align*} \Delta T_{\mathrm{f}} &= i m K_{\mathrm{f}} \\[1.5ex] &= (1)(0.3334~m)(4.133~^{\circ}\mathrm{C}~m^{-1}) \\[1.5ex] &= 1.3779~^{\circ}\mathrm{C} \end{align*}\]
Again, we have solved for the change in freezing point and numerically we have obtained a positive value but we know that the freezing point decreases so we will report a negative value.
A solution contains 350.0 g of a nonelectrolyte dissolved in 1.1 L of water The solution was found to boil at 104 °C. What is the molar mass of the solute? Kb(H2O) = 0.512 °C m–1
- 22.4
- 30.8
- 40.7
- 68.3
- 104.5
Solution
Answer: C
Concept: Boiling Point Elevation
We can use the boiling point elevation of a solution to determine the molar mass of the solute. The change in boiling point is 4.0 °C since water boils at 100 °C.
Find Molal Concentration
\[\begin{align*} \Delta T_{\mathrm{b}} &= i m K_{\mathrm{b}} \longrightarrow \\[1.5ex] m &= \dfrac{\Delta T_{\mathrm{b}}}{iK_{\mathrm{b}}} \\[1.5ex] &= \dfrac{4~^{\circ}\mathrm{C}}{(1)(0.512~^{\circ}\mathrm{C}~m^{-1})} \\[1.5ex] &= 7.8125~\mathrm{mol~kg^{-1}} \end{align*}\]
Find Moles of Solute
Since no density for water was provided, we assume 1.0 g mL–1.
\[\begin{align*} m &= \dfrac{n_{\mathrm{solute}}}{\mathrm{kg~solvent}} \longrightarrow \\[1.5ex] n_{\mathrm{solute}} &= m \times \mathrm{kg~solvent} \\[1.5ex] &= \left ( 7.8125~\mathrm{mol~kg^{-1}} \right ) \left ( 1.1~\mathrm{L} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{1~\mathrm{L}} \right ) \left ( \dfrac{1~\mathrm{g}}{1~\mathrm{mL}} \right ) \left ( \dfrac{1~\mathrm{kg}}{10^3~\mathrm{g}} \right ) \\[1.5ex] &= 8.59375~\mathrm{mol} \end{align*}\]
Find Molar Mass of Solute
\[\begin{align*} \mathrm{molar~mass} &= \dfrac{m}{n}\\[1.5ex] &= \dfrac{350.0~\mathrm{g}}{8.59375~\mathrm{mol}} \\[1.5ex] &= 40.727~\mathrm{g~mol^{-1}} \end{align*}\]
- 22.4
Choose the false statement regarding a 1 L aqueous solution that contains 1.0 g of NaCl. If all statements are true, choose “none of these are false”.
- χsolvent > χsolute
- Tf(aq. solution) > Tf(H2O)
- mol % solute < mol % solvent
- Tb(aq. solution) > Tb(H2O)
- none of these are false
Solution
Answer: B
The freezing point of a solution should be lower than that of a pure solvent.
150 g of KCl (m.m. = 74.5513 g mol–1) is completely dissolved in enough water to make a 834 mL solution (at 25 °C). What is the osmotic pressure (in atm) of the solution?
- 109.1
- 118.0
- 178.3
- 218.4
- 262.5
Solution
Answer: B
Concept: Osmotic Pressure
Osmotic pressure is given as
\[\Pi = iMRT\]
and to find it, we must first determine the molar concentration of the solution.
Find Moles of Solute
\[\begin{align*} n_{\mathrm{solute}} &= \dfrac{m}{\mathrm{molar~mass}} \\[1.5ex] &= \dfrac{150.0~\mathrm{g}}{74.5513~\mathrm{g~mol^{-1}}} \\[1.5ex] &= 2.012~\mathrm{mol} \end{align*}\]
Find Molar Concentration
\[\begin{align*} M &= \dfrac{\mathrm{mol~solute}}{\mathrm{L~solution}} \\[1.5ex] &= \dfrac{2.012~\mathrm{mol}}{834~\mathrm{mL} \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right )} \\[1.5ex] &= 2.41247~\mathrm{mol~L^{-1}} \end{align*}\]
Find Temperature in Kelvin
\[\begin{align*} T &= 273.15 + 25~^{\circ}\mathrm{C} \\[1.5ex] &= 298.15~\mathrm{K} \end{align*}\]
Find Osmotic Pressure
The van’t Hoff factor for KCl is 2.
\[\begin{align*} \Pi &= iMRT \\[1.5ex] &= (2)(2.41247~\mathrm{mol~L^{-1}})(0.082057~\mathrm{L~atm~mol^{-1}~K^{-1}}) (298.15~\mathrm{K})\\[1.5ex] &= 118.043~\mathrm{atm} \end{align*}\]