The vapour pressure of an aqueous solution is found to be 750 torr at a temperature, , and the same — Solutions and Colligative Properties Chemistry Question
Question
The vapour pressure of an aqueous solution is found to be 750 torr at a temperature, $T$, and the same solution shows an elevation in boiling point equal to 1.04 K. If $T$ is the boiling point of pure water, then the atmospheric pressure should be ($K_b \text{ of water} = 0.52 \text{ K-kg/mol}$)
💡 Solution & Explanation
$\Delta T_b = 1.04$ K. Molality $m = 1.04 / 0.52 = 2.0$ m. From $(P^\circ - P) / P = n / N$, we know $n/N = m \times M_{solvent} / 1000 = 2.0 \times 18 / 1000 = 0.036$. Substituting $P = 750$ torr, $(P^\circ - 750) / 750 = 0.036 \Rightarrow P^\circ = 750 + (750 \times 0.036) = 750 + 27 = 777$ torr. Since $T$ is the boiling point of pure water, the external atmospheric pressure equals the vapour pressure of pure water at $T$, which is 777 torr. Therefore, correct answer is B.