Question

At T(K), the vapor pressure of x molal aqueous solution containing a non-volatile solute is 12.078 kPa. The vapor pressure of pure water at T(K) is 12.3 kPa. What is the value of x?

• A. 10
• B. 1.018
• C. 0.1018
• D. 0.018

Hint:

Raoult's Law: $\frac{P_0 - P}{P_0} = \chi_{\text{solute}}$. For dilute solutions, $\chi_{\text{solute}} \approx \frac{x}{\frac{1000}{18}}$.

Answer 1

The Correct Option is B

Raoult's law states that the relative lowering of vapor pressure is equal to the mole fraction of the solute. $$ \frac{P_0 - P}{P_0} = \chi_{\text{solute}} $$ Where $P_0$ is the vapor pressure of pure solvent, $P$ is the vapor pressure of the solution, and $\chi_{\text{solute}}$ is the mole fraction of the solute. $$ \frac{12.3 - 12.078}{12.3} = \chi_{\text{solute}} $$ $$ \chi_{\text{solute}} = \frac{0.222}{12.3} \approx 0.018 $$ For dilute solutions, molality ($x$) is related to mole fraction by: $$ \chi_{\text{solute}} = \frac{x}{x + \frac{1000}{18}} \approx \frac{x}{\frac{1000}{18}} \quad (\text{since } x \ll \frac{1000}{18}) $$ $$ 0.018 = \frac{x}{55.56} $$ $$ x \approx 0.018 \times 55.56 \approx 1.00 $$ However, if we use the more precise formula (without approximation), $0.018 = x/(x + 55.56)$ which gives $x = 1.018$