Question

Consider the following arrangement in which a solution containing 20 g of haemoglobin in $1d{m}^3$ of the solution is placed in a right-hand compartment and pure water in a left-hand compartment. Both the compartments are separated by a semipermeable membrane. At equilibrium, the height of water in the right-hand column is 77.8 mm in excess of that in the left-hand column at 298 K. Calculate the molar mass of haemoglobin if $g=981cm/s^2$ and density of right-hand column be assumed $1g/c{m}^3$.

• A. $3.24\times {10}^{7}gmo{l}^{-1}$
• B. $6.49\times {10}^{7}gmo{l}^{-1}$
• C. $7.2\times {10}^{8}gmo{l}^{-1}$
• D. $1.44\times {10}^{8}gmo{l}^{-1}$

Answer 1

The correct option is B $6.49\times {10}^{7}gmo{l}^{-1}$

Let M g/mol be the molar mass of haemoglobin.

Number of moles of haemoglobin $=\frac{20g}{Mg/mol}=\frac{20}{M}mole$. The volume of the solution is 1 $d{m}^3$ or 1 L.

The molarity of the haemoglobin solution is $C=\frac{\frac{20}{M}mol}{1L}=\frac{20}{M}M$

The osmotic pressure $\Pi =\frac{7.78cm\times 1g/c{m}^3\times 981m/s^2}{76.0cm\times 13.595g/c{m}^3\times 981m/s^2}=0.00753$ atm.

The osmotic pressure $\Pi =CRT$

$0.00753atm=\frac{20g}{M}mol/L\times 0.08206Latm/molK\times 298K$

$M=6.49\times {10}^{4}g/mol$