Question

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

• A. 2
• B. 0.83
• C. 1
• D. 1.83

Answer 1

Answer: (B)

Given,

Concentration of $NaCl(C)=0.1M$

Mass of urea $\left(w_B\right)=1.1g$

Molar mass of urea $\left(M_B\right)=60gmo{l}^{-1}$

Molar mass of $NaCl\left(M_A\right)=58.5gmo{l}^{-1}$

Volume of solution $(V)=100mL$

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

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

$i\times CRT(NaCl)=CRT$ (urea)

where,

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

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

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

$i\times \frac{0.1}{1L}=\frac{w_B}{m_B}\times \frac{1000}{V}$

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

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 $(\alpha )-0.83$