Question

Relative decrease in vapour pressure of an aqueous solution containing $2$ mol $[Cu(NH)_{3})_{3}Cl] Cl$ in $3$ mol $H_{2}O$ is $\frac{1}{2}$ When the given solution reacts with excess of $AgNO_{3}$ solution the number of moles of $AgCl$ produced is

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

Answer 1

Answer: (A)

Let degree of ionisation of the complex, $[Cu(NH_{3})_{B}Cl]Cl]Cl$ be $\alpha$
$[Cu(NH_{3})_{3}Cl] \rightleftharpoons [Cu(NH_{3})Cl]^{+}+Cl^{-}$
$i=1+a$
$\frac{\Delta P}{P_{1}^{0}}=\frac{n_{1}\left(1+\alpha\right)}{n_{1}\left(1+\alpha\right)+n_{2}}$
$=\frac{2\left(1+\alpha\right)}{2\left(1+\alpha\right)+3}=\frac{1}{2}$
$\alpha=0.5$
Total $Cl^{-}$ produced $2\,\alpha=2\times\frac{1}{2}=1$ mole
$\therefore $ 1 mole of $AgCl$ is produced