Question

A $0.001$ molal solution of $\left[Pt \left(\left(NH\right)_{3}\right)_{4} \left(Cl\right)_{4}\right]$ in water has a freezing point depression of $0.0054^\circ C$ . If $K_{f}$ of water is $1.80$ , then correct formulation of above molecule is :

• A. $\left[Pt \left(\left(NH\right)_{3}\right)_{4} \left(Cl\right)_{3}\right]Cl$
• B. $\left[Pt \left(\left(NH\right)_{3}\right)_{4} \left(Cl\right)_{2}\right]\left(Cl\right)_{2}$
• C. $\left[Pt \left(\left(NH\right)_{3}\right)_{4} Cl\right]\left(Cl\right)_{3}$
• D. $\left[Pt \left(\left(NH\right)_{3}\right)_{4} \left(Cl\right)_{4}\right]$

Answer 1

Answer: (B)

$ΔT_{f}=iK_{f}\cdot m$
$0.0054=i\times 1.80\times 0.001$
$i=3$ if $\alpha =100\%$ then, $i=n$
$n=3$