Vapour pressure of a solution of 5 g of non-electrolyte in 100 g of water at a particular temperatur — Solutions and Colligative Properties Chemistry Question
Question
Vapour pressure of a solution of 5 g of non-electrolyte in 100 g of water at a particular temperature is $2985 \text{ N/m}^2$. If the vapour pressure of pure water at this temperature is $3000 \text{ N/m}^2$, the molecular mass of the solute is
Answer: C
💡 Solution & Explanation
$\Delta P = 3000 - 2985 = 15 \text{ N/m}^2$. Using $(P^\circ - P) / P = n / N$, we get $15 / 2985 = (5 / M) / (100 / 18)$. $15 / 2985 = (5 \times 18) / (100 \times M) = 90 / 100M = 0.9 / M$. Solving for M gives $M = (2985 \times 0.9) / 15 = 199 \times 0.9 = 179.1 \approx 180 \text{ g/mol}$. Therefore, correct answer is C.
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