Question

The vapor pressure of water at room temperature is lowered by $5\%$ by dissolving a solute in it. What is the approximate molality of the solution?

• A. $2$
• B. $1$
• C. $4$
• D. $3$

Answer 1

The correct option is D

$3$

The explanation for the correct option:

(D) 3

Step 1: Given information

• Initial vapor pressure of pure solvent (${\mathrm{P}}^0$) $=100\%$
• Vapor pressure after lowering ($\mathrm{P}$) $=100-5$

$=95\%$

Step 2: Formula used

• According to Raoult’s law, a solvent’s partial vapor pressure in a solution is equal to the vapor pressure of the pure solvent multiplied by its mole fraction in the solution.
• Raoult’s law equation is written as;

${\mathrm{P}}_{\left(\mathrm{solution}\right)}={\mathrm{X}}_{\left(\mathrm{solvent}\right)}{\mathrm{P}}_{\left(\mathrm{solvent}\right)}$

• Relation of vapor pressure with molality;

$\frac{{\mathrm{P}}^0-\mathrm{P}}{\mathrm{P}}=\frac{\mathrm{m}\times \mathrm{M}}{1000}$

Where,

${\mathrm{P}}^0$ is the vapor pressure of the pure solvent

$\mathrm{P}$ is the vapor pressure after lowering

$\mathrm{m}$ is the molality

$\mathrm{M}$ is the molar mass

Step 3: Calculation

• $\frac{{\mathrm{P}}^0-\mathrm{P}}{\mathrm{P}}=\frac{\mathrm{m}\times \mathrm{M}}{1000}$
• $\frac{100-95}{95}$ $=$ $\frac{\mathrm{m}\times 18}{1000}$
• $\mathrm{m}=3$

The explanation for the incorrect options:

Since, the correct value is 3. Therefore, the options (A), (B) and (C) stand incorrect.

Therefore, the correct answer is option (D) 3.