Question

Relative decrease in vapour pressure of an aqueous solution containing $2$ moles $\left[Cu{\left({NH}_3\right)}_{3}Cl\right]Cl$ in $3$ moles $H_{2}O$ is $0.50$. On reaction with $Ag{NO}_3$, this solution will form:

• A. $1$ mol $AgCl$
• B. $0.25$ mol $AgCl$
• C. $2$ mol $AgCl$
• D. $0.40$ mol $AgCl$

Answer 1

Answer: (A)

$1$ mol $AgCl$

Relative lowering in vapour pressure= $0.50$

$ix=0.5$

$x$= mole fraction

As, $x=\frac{2}{5}=\frac{molesofsolute}{molesofsolution}$

$i=0.5\times \frac{5}{2}=1.25$

Now, degree of dissociation

$\alpha =\frac{i-1}{n-1}=\frac{1.25-1}{2-1}=0.25$

$(n=2\therefore ions=2)$

So, moles of chlorine react with $AgN{O}_3=1$ and Moles of $AgCl$ precipitate= $1\times 0.25=0.25mol$