20 g hemoglobin in a 1 L aqueous solution (A) at 300 K is separated from pure water by a semi-permeable membrane. At equilibrium, the height of solution in a tube dipped in solution (A) is found to be $$80.0 mm$$ higher than the tube dipped in water.
The molar mass of hemoglobin is ______ $$kg mol^{-1}$$. (Nearest integer)
(Given: $$g = 10 \, m \, s^{-2}$$, $$R = 8.3 \, kPa \, dm^{3} \, K^{-1} \, mol^{-1}$$, density of solution = $$1000 \, kg \, m^{-3}$$)

Given:

• Mass of hemoglobin $$w=20 g$$
• Volume of solution $$V=1 L=10^{−3} m^{3}$$
• Temperature $$T=300 K$$
• Height difference $$h=80.0 mm =0.08 m$$
• Density $$ρ=1000 kg m^{−3}$$ 
• $$g=10 m s^{−2}$$
• $$R=8.3 kPa dm^{3}K^{−1}mol^{−1}$$

Osmotic Pressure$$\left(\pi\right)$$:

$$\pi=\rho gh$$

$$=1000\times\ 10\times\ 0.08$$

$$800Pa=0.8kPa$$

Since, $$\pi\ V=nRT$$

and $$n=\frac{w}{M}$$

so,

$$\pi\ V=\frac{w}{M}RT$$

$$M=\frac{wRT}{\pi\ V}$$

$$M=\frac{\left(20\times\ 8.3\times\ 300\right)}{08\times\ 1}$$

$$M=\frac{49800}{0.8}$$

$$M=62250gmol^{-1}$$

$$M=62.25kgmol^{-1}$$

Therefore, nearest integer is 62.