Question

A quantity of $10 \,gm$ of solute $'A'$ and $20\, g$ of solute '$B$ ' is dissolved in $500 \,ml$ water. The solution is isotonic with the solution obtained by dissolving $6.67\,g$ of 'A' and $30\, g$ of ' $B$ ' in $500 ml$ 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$'A'+20 g$ of solute ' $B$ ' dissolved in 500 $ml H _{2} O$

$M=\frac{\frac{10}{M_{A}}+\frac{20}{M_{B}}}{\frac{500}{100}}$

$=\left(\frac{10}{M_{A}}+\frac{20}{M_{B}}\right) \times \frac{1000}{500}$

$M=\frac{20}{M_{A}}+\frac{40}{M_{B}}=\frac{20 M_{B}+40 M_{A}}{M_{A} M_{B}}$

Concentration in Case II

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

$M=\frac{\frac{6.67}{M_{A}}+\frac{30}{M_{B}}}{\frac{500}{1000}}$

$=\left(\frac{6.67}{M_{A}}+\frac{30}{M_{B}}\right) \times \frac{1000}{500}$

$M=\frac{13.34}{M_{A}}+\frac{60}{M_{B}}$

$=\frac{13.34 M_{B}+60 M_{A}}{M_{A} M_{B}}$

Since solutions are isotonic, concentrations must be same:

$\therefore \frac{20 M _{ B }+40 M _{ A }}{ M _{ A } M _{ B }}=\frac{13.34 M _{ B }+60 M _{ A }}{ M _{ A } M _{ B }}$

$20 M _{ B }-13.34 M _{ B }=60 M _{ A }-40 M _{ A }$

$6.66 M _{ B }=20 M _{ A }$

$\frac{1}{3}=\frac{6.66}{20}=\frac{ M _{ A }}{ M _{ B }}$