Question

If $0.1\, M$ solution of $NaCl$ is isotonic with $1.1\, w \%$ urea solution, the degree of ionisation of $NaCl$ is (molar masses of urea and $NaCl$ are $60$ and $58.5\, g\, mol ^{-1}$ respectively $)$

• A. 2
• B. 0.83
• C. 1
• D. 1.83

Answer 1

Answer: (B)

Given,

Concentration of $NaCl(C)=0.1\, M$

Mass of urea $\left(w_{B}\right)=1.1\, g$

Molar mass of urea $\left(M_{B}\right)=60\, g\, mol ^{-1}$

Molar mass of $NaCl \left(M_{A}\right)=58.5\, g\, mol ^{-1}$

Volume of solution $(V)=100\, mL$

$\because$ Solutions of $NaCl$ and urea are isotonic.

$\therefore \pi( NaCl )=\pi($ urea $)$

$i \times C R T( NaCl )=C R T$ (urea)

where,

$i=$ van't Hoff factor, $C=$ concentration

$R=$ gas constant, $T= $ temperature

$\therefore i \times C( NaCl )=C(\text { urea })$

$i \times \frac{0.1}{1\, L } =\frac{w_{B}}{m_{B}} \times \frac{1000}{V}$

$i=\frac{1.1}{60} \times \frac{1000}{100} \times \frac{1}{0.1}=1.83$

Also, for ionisation $(i)-1-\alpha+n \alpha$

where, $\alpha=$ degree of dissociation

$n=$ number of particles $=2($ for $NaCl )$

Thus,

$i =1-\alpha+2 \alpha$

$1.83 =1+\alpha$

or, $\alpha =1.83-1.0$

Degree of dissociation $(\alpha)-0.83$