Consider the following arrangement, in which a solution containing 20 g of haemoglobin in of the sol — Solutions and Colligative Properties Chemistry Question
Question
Consider the following arrangement, in which a solution containing 20 g of haemoglobin in $1 \text{ dm}^3$ of the solution is placed in right compartment and pure water is placed in left compartment, separated by SPM. At equilibrium, the height of liquid in the right compartment is 74.5 mm in excess of that in the left compartment. The temperature of the system is maintained at 298 K. The number of millimoles in 320 g of haemoglobin 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 $\pi = \rho g h = 1013 \text{ kg/m}^3 \times 10 \text{ m/s}^2 \times 0.0745 \text{ m} = 754.685 \text{ Pa}$. In atm, $\pi = 754.685 / 101325 \approx 0.007448 \text{ atm}$. Using $\pi = CRT = (w / (M \cdot V)) R T$, we get $0.007448 = (20 / (M \times 1)) \times 0.0821 \times 298$. $M = (20 \times 24.4658) / 0.007448 \approx 65697 \text{ g/mol}$. Moles in 320 g = $320 / 65697 \approx 0.00487 \text{ mol} = 4.87 \text{ millimoles}$. Rounding to the nearest integer gives 5. Therefore, correct answer is 5.