The vapour pressure of pure solvent A is . At the same temperature, when of a non-volatile solute B — Solutions and Colligative Properties Chemistry Question
Question
The vapour pressure of pure solvent A is $10\text{ torr}$. At the same temperature, when $1\text{ g}$ of a non-volatile solute B is dissolved in $20\text{ g}$ of A, its vapour pressure is reduced to $9.0\text{ torr}$. If the molecular mass of A is $200\text{ amu}$, the molecular mass of B is:
Answer: B
💡 Solution & Explanation
Using the exact RLVP formula: $\frac{P^0_A - P_S}{P_S} = \frac{n_B}{n_A}$. We have $\frac{10 - 9}{9} = \frac{1/M_B}{20/200}$. This yields $\frac{1}{9} = \frac{1/M_B}{0.1} = \frac{10}{M_B}$. Solving for $M_B$ gives $M_B = 90\text{ amu}$.
💬Ask on WhatsApp →
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes