Question

$1 \times 10^{-3} m$ solution of $Pt \left( NH _{3}\right)_{2} Cl _{4}$ in $H _{2} O$ shows depression in freezing point by $0.0054^{\circ} C$. The ionisable $Cl ^{-}$ions will be (Given, $K_{f}\left( H _{2} O \right)=1.860\, K\, kg\, mol ^{-1}$ )

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (B)

$\Delta T_{f}=i K_{f} m$
$\Rightarrow 0.0054=i \times 1.86 \times 0.001$
$i=\frac{5.4}{1.86} \approx 3$
$\left[ Pt \left( NH _{3}\right)_{2} Cl _{2}\right] Cl _{2} \rightleftharpoons\left[ Pt \left( NH _{3}\right)_{2} Cl _{2}\right]^{2+}+2 Cl ^{-}$
So, ionisable $Cl ^{-}$ions are $2$ .