Question

What is vapour pressure of a solution containing 0.1 mol of non-volatile solute dissolved in 16.2 g of water ? ($P_1^0= 32\text{ mm Hg}$)

• A. $21.6\text{ mm Hg}$
• B. $28.8\text{ mm Hg}$
• C. $15.7\text{ mm Hg}$
• D. $18.1\text{ mm Hg}$

Hint:

Always ensure you are calculating the mole fraction of the solvent ($x_1$) when using $P = P^0 \cdot x_1$, not the solute! If you calculate the mole fraction of the solute ($x_2$), use relative lowering: $\frac{P^0 - P}{P^0} = x_2$.

Answer 1

The Correct Option is B

Step 1: Use Raoult's law.
For a non-volatile solute, $P_1 = P_1^0 \times x_1$, where $x_1$ is the mole fraction of water.

Step 2: Count the moles.
Solute is $0.1$ mol. Water is $\dfrac{16.2}{18} = 0.9$ mol. So $x_1 = \dfrac{0.9}{0.9 + 0.1} = 0.9$.

Step 3: Find the vapour pressure.
$$P_1 = 32 \times 0.9 = 28.8\ \text{mm Hg}$$
\[ \boxed{28.8\ \text{mm Hg}} \]