Solve the following

Question:

If $250 \mathrm{~cm}^{3}$ of an aqueous solution containing $0.73 \mathrm{~g}$ of a protein $\mathrm{A}$ is isotonic with one litre of another aqueous solution containing $1.65 \mathrm{~g}$ of a protein $\mathrm{B}$, at $298 \mathrm{~K}$, the ratio of the molecular masses of $\mathrm{A}$ and $\mathrm{B}$ is______________ $\times 10^{-2}$ (to the nearest integer). 

Solution:

(177)

$\pi_{A}=i C_{A} R T, \pi_{B}=i C_{B} R T$

For isotonic solution, $\pi_{A}=\pi_{B}$

$i_{1} C_{1}=i_{2} C_{2}$            (For protein $i=1$ )

$\Rightarrow C_{1}=C_{2}$

$\Rightarrow \frac{0.73 \times 1000}{M_{A} \times 250}=\frac{1.65}{M_{B} \times 1}$

$\therefore \frac{M_{A}}{M_{B}}=\frac{0.73 \times 4}{1.65}=1.77=177 \times 10^{-2}$

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