Question

35 g of $\text{N}{\text{H}}_{\text{4}}\text{OH}$ is dissolved in 10L of ${\text{H}}_{\text{2}}\text{O}$. The percentage dissociation of $\text{N}{\text{H}}_{\text{4}}\text{OH}$ is-
$\left(\text{M}\text{.wt}\text{N}{\text{H}}_{\text{4}}\text{OH - 35;}\text{pK,N}{\text{H}}_{\text{4}}\text{OH - 4}\text{.75}\right)$

• A. 13.4
• B. 1.34
• C. 26.8
• D. 2.68

Answer 1

Answer: (B)

1.34

Given that

$pKb=4.75\Rightarrow K_b=1.8\times {10}^{-5}$

$N{H}_{4}OH=35g$

Concentration $C$ of $N{H}_{4}OH=\frac{35}{35\times 10}$

$=0.1M$

We know that,

$\alpha =\sqrt{\frac{K_b}{c}}=\sqrt{\frac{1.8\times {10}^{-5}}{0.1}}$

$\alpha =1.34\times {10}^{-2}$

$\%\alpha =1.34$

Option -(B)