Q16 / 180 · COMEDK 2025 (10 May Shift 2) · Chemistry
An aqueous solution which contains $42 \%$ by weight ( $\mathrm{w} / \mathrm{W}$ ) of a volatile liquid " A " of molar mass of $140 \mathrm{~g} / \mathrm{mol}$, has a vapour pressure of 2141.4 mm of Hg at $37^{\circ} \mathrm{C}$. What is the vapour pressure of the pure liquid "A"? Vapour pressure of water at $37^{\circ} \mathrm{C}$ is 2019.1 mm .
- A293 nm
- B2519.2 nm
- Correct answerC3455.0 nm
- D2356.4 nm
Solution
Let the total mass of the solution be $100 \text{ g}$.
Mass of solute A = $42 \text{ g}$.
Mass of solvent (water) = $100 - 42 = 58 \text{ g}$.
Molar mass of A ($M_A$) = $140 \text{ g/mol}$.
Molar mass of water ($M_W$) = $18 \text{ g/mol}$.
Number of moles of A ($n_A$) = $\dfrac{42}{140} = 0.3 \text{ mol}$.
Number of moles of water ($n_W$) = $\dfrac{58}{18} = 3.222 \text{ mol}$.
Mole fraction of A ($x_A$) = $\dfrac{n_A}{n_A + n_W} = \dfrac{0.3}{0.3 + 3.222} = \dfrac{0.3}{3.522} \approx 0.08518$.
Mole fraction of water ($x_W$) = $1 - x_A = 1 - 0.08518 = 0.91482$.
According to Raoult's Law for a solution of two volatile liquids:
$P_{total} = P_A^{0} x_A + P_W^{0} x_W$
Given $P_{total} = 2141.4 \text{ mm Hg}$ and $P_W^{0} = 2019.1 \text{ mm Hg}$:
$2141.4 = P_A^{0} (0.08518) + 2019.1 (0.91482)$
$2141.4 = P_A^{0} (0.08518) + 1847.10$
$P_A^{0} (0.08518) = 2141.4 - 1847.10 = 294.3$
$P_A^{0} = \dfrac{294.3}{0.08518} \approx 3455.03 \text{ mm Hg}$..
Answer: 3455.0 nm
Mass of solute A = $42 \text{ g}$.
Mass of solvent (water) = $100 - 42 = 58 \text{ g}$.
Molar mass of A ($M_A$) = $140 \text{ g/mol}$.
Molar mass of water ($M_W$) = $18 \text{ g/mol}$.
Number of moles of A ($n_A$) = $\dfrac{42}{140} = 0.3 \text{ mol}$.
Number of moles of water ($n_W$) = $\dfrac{58}{18} = 3.222 \text{ mol}$.
Mole fraction of A ($x_A$) = $\dfrac{n_A}{n_A + n_W} = \dfrac{0.3}{0.3 + 3.222} = \dfrac{0.3}{3.522} \approx 0.08518$.
Mole fraction of water ($x_W$) = $1 - x_A = 1 - 0.08518 = 0.91482$.
According to Raoult's Law for a solution of two volatile liquids:
$P_{total} = P_A^{0} x_A + P_W^{0} x_W$
Given $P_{total} = 2141.4 \text{ mm Hg}$ and $P_W^{0} = 2019.1 \text{ mm Hg}$:
$2141.4 = P_A^{0} (0.08518) + 2019.1 (0.91482)$
$2141.4 = P_A^{0} (0.08518) + 1847.10$
$P_A^{0} (0.08518) = 2141.4 - 1847.10 = 294.3$
$P_A^{0} = \dfrac{294.3}{0.08518} \approx 3455.03 \text{ mm Hg}$..
Answer: 3455.0 nm
Hence, the correct answer is option: 'C'.