Question

The vapour pressure of water is $12.3kPa$ at $300K$. What is the vapour pressure of 1 molal aqueous solution of a non-volatile solute at $300K?$

• A. 1.208 kPa
• B. 12.08 kPa
• C. 1208 kPa
• D. 2.08 kPa

Answer 1

Answer: (B)

$\frac{P^0-P_S}{P_S}=\frac{n_B}{n_A}$

1 molal means 1 mole of solute is dissolved in $1kg$ of solvent (water here)

$\frac{12.3-P_S}{12.3}=\frac{1}{1+\frac{1000}{18}}=\frac{1}{1+55.56}$

$\frac{12.3-P_S}{12.3}=0.0177$

$12.3-P_s=0.2177$

$P_s=12.08kPa$