Question

Relative decrease in vapour pressure of an aqueous solution containing 2 moles $\left[Cu\right(N{H}_3{)}_{3}Cl]$ Cl in 3 moles $H_{2}O$ is 0.50. On reaction with $AgN{O}_3$, this solution will form:

[assuming no change in degree of ionisation of substance on adding $AgN{O}_3$]

• A. 1 mol $AgCl$
• B. 0.25 mol $AgCl$
• C. 0.5 mol $AgCl$
• D. 0.40 mol $AgCl$

Answer 1

The correct option is A 1 mol $AgCl$

The mole fraction is $\frac{i.n}{i.n+N}=\frac{i.2}{i.2+3}$

The relative lowering in vapour pressure is

$\frac{\Delta P}{P}=0.5=\frac{1}{2}$

But $\frac{\Delta P}{P}=\text{mole fraction}$

Hence, $\frac{1}{2}=\frac{i.2}{i.2+3}$

$i=\frac{3}{2}=1.5$

$1.5=1+(n-1)\alpha$

$\alpha =\frac{1}{2}$

The number of moles of $C{l}^{-}ions=2\times \frac{1}{2}=1moles$.

Hence 1 mol of $AgCl$ is precipitated.

Hence, the correct option is $A$