Question

If 250 cm³ of an aqueous solution containing 0.73 g of protein A is isotonic with one liter 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 ______ × 10^–2 (to the nearest integer).

Answer 1

Step 1: Given information

$$\begin{gathered} {\mathrm{m}}_1=0.73\mathrm{g} \\ {\mathrm{v}}_1=250\mathrm{mL}=0.25\mathrm{L} \\ {\mathrm{m}}_2=1.65\mathrm{g} \\ {\mathrm{v}}_2=1\mathrm{L} \end{gathered}$$

As we know $\pi v=nRT$

$\pi =$Osmotic pressure, v= number of moles, R= gas constant, and T= temperature, Isotonic = concentration same

$\begin{array}{l}\therefore \pi =\frac{n}{v}RT\\ \Rightarrow \pi =CRT\end{array}$

Step 2: According to the question ${\mathrm{C}}_1(\mathrm{Protein}\mathrm{A})={\mathrm{C}}_2(\mathrm{Protein}\mathrm{B})$

$$\begin{gathered} \text{Therefore,}\frac{{\mathrm{n}}_1}{{\mathrm{v}}_1}=\frac{{\mathrm{n}}_2}{{\mathrm{v}}_2} \\ \Rightarrow \frac{{\displaystyle \raisebox{1ex}{${\mathrm{m}}_1$}\!\left/ \!\raisebox{-1ex}{${\mathrm{M}}_1$}\right.}}{{\mathrm{v}}_1\left(\mathrm{litre}\right)}=\frac{{\displaystyle \raisebox{1ex}{${\mathrm{m}}_2$}\!\left/ \!\raisebox{-1ex}{$\mathrm{M}$}\right.}}{{\mathrm{v}}_2\left(\mathrm{litre}\right)}, \\ \Rightarrow \frac{{\displaystyle \raisebox{1ex}{$0.73$}\!\left/ \!\raisebox{-1ex}{${\mathrm{M}}_1$}\right.}}{0.25}=\frac{{\displaystyle \raisebox{1ex}{$1.65$}\!\left/ \!\raisebox{-1ex}{${\mathrm{M}}_2$}\right.}}{1} \\ \Rightarrow \frac{0.73}{0.25\times {\mathrm{M}}_1}=\frac{1.65}{1\times {\mathrm{M}}_2} \\ \Rightarrow \frac{0.73}{0.25\times 1.65}=\frac{{\mathrm{M}}_1}{{\mathrm{M}}_2} \\ \Rightarrow 0.0176=\frac{{\mathrm{M}}_1}{{\mathrm{M}}_2} \\ \Rightarrow 1.76\times {10}^{-2}=\frac{{\mathrm{M}}_1}{{\mathrm{M}}_2} \\ \Rightarrow 2\times {10}^{-2}=2 \end{gathered}$$

Final Answer: Hence, the correct answer is 2.