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

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Let molar mass of protein A = x g/mol
Let molar mass of protein B = y g/mol

\pi_{\mathrm{A}}=\text { osmotic pressure of protein } \mathrm{A}=\frac{\left(\frac{0.73}{\mathrm{x}}\right)}{0.25} \mathrm{RT}

\pi_{\mathrm{B}}=\text { osmotic pressure of protein } \mathrm{B}=\frac{\left(\frac{1.65}{\mathrm{y}}\right)}{1} \mathrm{RT}

Now, \pi_{\mathrm{A}}=\pi_{\mathrm{B}}

\left(\frac{0.73}{\mathrm{x} \times 0.25}\right) \mathrm{RT}=\left(\frac{1.65}{\mathrm{y}}\right) \mathrm{RT}

\left(\frac{\mathrm{x}}{\mathrm{y}}\right)=\frac{0.73}{0.25 \times 1.65}=1.769 \cong 1.77

\mathrm{\frac{x}{y} = 177 \times 10^{-2}}

Ans = 177

Posted by

Kuldeep Maurya

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