Question

What is vapour pressure of solution containing $0.1\ \mathrm{mole}$ solute dissolved in $1.8 \times 10^{-2}\ \mathrm{kg}\ \mathrm{H_2O}$? ($P_1^\circ = 24\ \mathrm{mm\ Hg}$)

• A. $12.40\ \mathrm{mm\ Hg}$
• B. $18.12\ \mathrm{mm\ Hg}$
• C. $15.72\ \mathrm{mm\ Hg}$
• D. $21.84\ \mathrm{mm\ Hg}$

Hint:

When the mole fraction is small, a quick shortcut approximation is $\frac{P_1^\circ - P_1}{P_1^\circ} \approx \frac{n_2}{n_1}$. Here, $\frac{\Delta P}{24} \approx \frac{0.1}{1} = 0.1 \implies \Delta P \approx 2.4\ \mathrm{mm\ Hg}$. Subtracting $2.4$ from $24$ gives $21.6\ \mathrm{mm\ Hg}$, which easily points to the closest exact choice, $21.84\ \mathrm{mm\ Hg}$!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given a solution where $n_2 = 0.1\ \mathrm{mol}$ of solute is dissolved in a mass of water $W_1 = 1.8 \times 10^{-2}\ \mathrm{kg}$. Given the vapour pressure of pure water is $P_1^\circ = 24\ \mathrm{mm\ Hg}$, we need to calculate the vapour pressure of the resulting solution ($P_1$).

Step 2: Key Formula or Approach:
According to Raoult's law for a non-volatile solute, the relative lowering of vapour pressure is equal to the mole fraction of the solute ($x_2$): $$\frac{P_1^\circ - P_1}{P_1^\circ} = x_2 = \frac{n_2}{n_1 + n_2}$$ Where $n_1$ is the number of moles of the solvent ($\mathrm{H_2O}$), calculated via $n_1 = \frac{W_1}{M_1}$.

Step 3: Detailed Explanation:
First, convert the mass of water to grams to compute its moles: $$W_1 = 1.8 \times 10^{-2}\ \mathrm{kg} = 18\ \mathrm{g}$$ The molar mass of water ($\mathrm{H_2O}$) is $M_1 = 18\ \mathrm{g\ mol^{-1}}$. $$n_1 = \frac{18\ \mathrm{g}}{18\ \mathrm{g\ mol^{-1}}} = 1\ \mathrm{mol}$$ Now, find the mole fraction of the solute ($x_2$): $$x_2 = \frac{0.1}{1 + 0.1} = \frac{0.1}{1.1} = \frac{1}{11} \approx 0.0909$$ Substitute this mole fraction value into Raoult's equation: $$\frac{24 - P_1}{24} = \frac{1}{11}$$ $$24 - P_1 = \frac{24}{11} \approx 2.16\ \mathrm{mm\ Hg}$$ Isolate the vapour pressure of the solution ($P_1$): $$P_1 = 24 - 2.16 = 21.84\ \mathrm{mm\ Hg}$$

Step 4: Final Answer:
The vapour pressure of the solution is $21.84\ \mathrm{mm\ Hg}$, which corresponds to option (D).