If $250 \mathrm{~cm}^{3}$ of an aqueous solution containing $0.73 \mathrm{~g}$ of a protein A is isotonic with one litre of another aqueous solution containing $1.65 \mathrm{~g}$ of a protein B, at $298 \mathrm{~K}$, the ratio of the molecular mases of $\mathrm{A}$ and $\mathrm{B}$ is $\times 10^{-2}$ (to the nearest integer).
Let molar mass of protein $\mathrm{A}=\mathrm{x} \mathrm{g} / \mathrm{mol}$
Let molar mass of protein $\mathrm{B}=\mathrm{y} \mathrm{g} / \mathrm{mol}$
$\pi_{\mathrm{A}}=$ osmotic pressure of protein $\mathrm{A}=\frac{\left(\frac{0.73}{\mathrm{x}}\right)}{0.25} \mathrm{RT}$
$\pi_{\mathrm{B}}=$ osmotic pressure of protein $\mathrm{B}=\frac{\left(\frac{1.65}{\mathrm{y}}\right)}{1} \mathrm{RT}$
$\pi_{\mathrm{A}}=\pi_{\mathrm{B}}$
$\Rightarrow\left(\frac{0.73}{x \times 0.25}\right) R T=\left(\frac{1.65}{y}\right) R T$
$\Rightarrow\left(\frac{x}{y}\right)=\frac{0.73}{0.25 \times 1.65}=1.769 \cong 1.77$