Question

A quantity of $10gm$ of solute ${}^{\prime }{A}^{\prime }$ and $20g$ of solute '$B$ ' is dissolved in $500ml$ water. The solution is isotonic with the solution obtained by dissolving $6.67g$ of 'A' and $30g$ of ' $B$ ' in $500ml$ water at the same temperature. The ratio of molar masses $M_A:M_B$ is:

• A. 1: 1
• B. 3: 1
• C. 1: 3
• D. 2: 3

Answer 1

Answer: (C)

Concentration in case I

10 gms of solute${}^{\prime }{A}^{\prime }+20g$ of solute ' $B$ ' dissolved in 500 $ml{H}_{2}O$

$M=\frac{\frac{10}{M_A}+\frac{20}{M_B}}{\frac{500}{100}}$

$=\left(\frac{10}{M_A}+\frac{20}{M_B}\right)\times \frac{1000}{500}$

$M=\frac{20}{M_A}+\frac{40}{M_B}=\frac{20M_B+40M_A}{M_A{M}_B}$

Concentration in Case II

$6.67g$ of $A+30g$ of $B$ in $500ml{H}_{2}O$

$M=\frac{\frac{6.67}{M_A}+\frac{30}{M_B}}{\frac{500}{1000}}$

$=\left(\frac{6.67}{M_A}+\frac{30}{M_B}\right)\times \frac{1000}{500}$

$M=\frac{13.34}{M_A}+\frac{60}{M_B}$

$=\frac{13.34M_B+60M_A}{M_A{M}_B}$

Since solutions are isotonic, concentrations must be same:

$\therefore \frac{20M_B+40M_A}{M_A{M}_B}=\frac{13.34M_B+60M_A}{M_A{M}_B}$

$20M_B-13.34M_B=60M_A-40M_A$

$6.66M_B=20M_A$

$\frac{1}{3}=\frac{6.66}{20}=\frac{M_A}{M_B}$