Question

In 1 litre aqueous solution 20 g of Haemoglobin is present at 300 K. This solution has height difference of 80 mm, when separated from pure water through S.P.M. If density of solution is 1000 kg/m$^3$, then calculate molar mass of Haemoglobin in kg/mol. (Use : $g = 10$ m/s$^2$)

Hint:

For osmotic pressure problems with height difference, first calculate \(\pi\) using \(\pi = h\rho g\), then apply \(\pi = \frac{wRT}{MV}\) to find molar mass.

Answer 1

Correct Answer: 62

Step 1: Use the relation between osmotic pressure and hydrostatic pressure.
When the solution is separated from pure water through a semipermeable membrane, the osmotic pressure is balanced by the hydrostatic pressure due to height difference. Therefore,
\[ \pi = h \rho g \] Here,
\[ h = 80 \text{ mm} = 0.08 \text{ m}, \qquad \rho = 1000 \text{ kg/m}^3, \qquad g = 10 \text{ m/s}^2 \] So,
\[ \pi = 0.08 \times 1000 \times 10 = 800 \text{ Pa} \]
Step 2: Apply the osmotic pressure formula.
For a dilute solution, osmotic pressure is given by van't Hoff equation:
\[ \pi = \frac{n}{V}RT \] Since,
\[ n = \frac{w}{M} \] therefore,
\[ \pi = \frac{wRT}{MV} \] From this, molar mass is:
\[ M = \frac{wRT}{\pi V} \]
Step 3: Substitute the given values.
Given,
\[ w = 20 \text{ g} = 0.02 \text{ kg} \] \[ V = 1 \text{ litre} = 10^{-3} \text{ m}^3 \] \[ T = 300 \text{ K} \] \[ R = 8.314 \text{ J mol}^{-1}\text{K}^{-1} \] \[ \pi = 800 \text{ Pa} \] Now substitute in the formula:
\[ M = \frac{0.02 \times 8.314 \times 300}{800 \times 10^{-3}} \]
Step 4: Calculate the value of molar mass.
\[ M = \frac{49.884}{0.8} \] \[ M = 62.355 \text{ kg/mol} \] Thus, the molar mass of haemoglobin is approximately:
\[ M \approx 62 \text{ kg/mol} \]
Step 5: State the final answer.
Hence, the molar mass of haemoglobin is:
\[ \boxed{62 \text{ kg/mol}} \]