At , a solution containing 0.2 g of polyisobutylene in 100 ml of benzene, developed a rise of 2.463 — Solutions and Colligative Properties Chemistry Question
Question
At $27^\circ\text{C}$, a solution containing 0.2 g of polyisobutylene in 100 ml of benzene, developed a rise of 2.463 mm in osmotic equilibrium. The molar mass of polyisobutylene (in $10^5 \text{ g/mol}$) is (Given: The density of final solution is $1.013 \text{ g/ml}$, $g = 10 \text{ m/s}^2$, $R = 0.0821 \text{ L-atm/K-mol}$)
💡 Solution & Explanation
Osmotic pressure $\pi = \rho g h$. Density $\rho = 1.013 \text{ g/ml} = 1013 \text{ kg/m}^3$. Height $h = 2.463 \text{ mm} = 0.002463 \text{ m}$. $\pi = 1013 \times 10 \times 0.002463 \approx 24.95 \text{ Pa}$. Converting to atm (1 atm = 101325 Pa): $\pi \approx 2.462 \times 10^{-4} \text{ atm}$. Using $\pi = CRT = (w / (M \cdot V)) R T$, we have $2.462 \times 10^{-4} = (0.2 / (M \times 0.1)) \times 0.0821 \times 300 = (2 / M) \times 24.63$. Solving for $M$: $M = (2 \times 24.63) / (2.462 \times 10^{-4}) \approx 2 \times 10^5 \text{ g/mol}$. The value in terms of $10^5 \text{ g/mol}$ is 2. Therefore, correct answer is 2.