Question

A solution containing 500 g of a protein per litre is isotonic with a solution containing 3.42 g sucrose per litre. The molecular mass of protein is $5\times 10^{x}$ , hence x is

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (C)

In case of isotonic solution

$\pi _{1}=\pi _{2}$

$C_{1}=C_{2}$

$\left(\frac{w_{1}}{m_{1} v_{1}}\right)_{p r o t e i n}=\left(\frac{w_{2}}{m_{2} v_{2}}\right)_{s u c r o s e}$

$\frac{500}{5 \times 1 0^{x} \times 1}=\frac{3.42}{342 \times 1}$

$\frac{100}{1 0^{x}}=\frac{1}{100}$

$10^{x}=10^{4}$

$x=4$