Question

Which of the following solutions are isotonic with respect to $0.5M$ $NaCl$ solution having the degree of dissociation = 0.8.

• A. $153.9g$ of sucrose is dissolved in $500mL$ of solution
• B. $0.3M$ aq. $N{a}_{2}S{O}_4$ solution undergoing $100\%$ ionization
• C. $1.8\%$ w/w aq.$NaOH$ solution undergoing $100\%$ ionization
• D. $1.2M$ benzoic acid in a solution of benzene where the acid dimerizes to an extent of $50\%$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $153.9g$ of sucrose is dissolved in $500mL$ of solution
B $0.3M$ aq. $N{a}_{2}S{O}_4$ solution undergoing $100\%$ ionization
C $1.8\%$ w/w aq.$NaOH$ solution undergoing $100\%$ ionization
D $1.2M$ benzoic acid in a solution of benzene where the acid dimerizes to an extent of $50\%$

At particular temperature isotonic solutions are those which have same solute concentration.

For $NaCl$:
$ic=0.5\times [1+(2-1\left)0.8\right]=0.9M$

For sucrose solution:
$ic=\frac{153.9}{342}\times \frac{1000}{500}\times 1=0.9M$

For $N{a}_{2}S{O}_4$ solution:
$ic=0.3\times 3=0.9M$

For $NaOH$:
Molality =$\frac{1.8}{40}\times \frac{1000}{100}$

For very dilute solutions molality is almost equal to the molarity. Thus
ic $=\frac{1.8}{40}\times 10\times 2=0.9M$

For benzoic acid:
$ic=1.2\times [1+(0.5-1\left)0.5\right]=1.2\times \frac{3}{4}=0.9M$

Solutions having same $ic$ value are isotonic.