0.1 M KI and 0.2M AgNO₃ are mixed in 3:1 volume ratio. The depression of freezing point of the resulting solution will be
(K_f(H₂O) = 1.86 K Kg mol–1)

 $$\begin{gathered} \mathrm{KI} + {\mathrm{AgNO}}_3\to {\mathrm{KNO}}_3+\mathrm{AgI}(\mathrm{ppt}.) \\ \mathrm{given}: 0.1\mathrm{M},3\mathrm{L} 0.2\mathrm{M}.1\mathrm{L} 0 \\ \mathrm{n}\left(\mathrm{MV}\right): 0.3 0.2 0.2\mathrm{moles} \end{gathered}$$

0.1 mole excess of KI is present

Total moles of solute ions are from Excess of KI and from KNO₃

Total no. of ions= (0.1+0.2)2=0.6moles (each contains 2 ions)

Total vol. of the sol. = (3+1)l =4lit

$∆{\mathrm{T}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{f}}\left[\mathrm{solute}\right]$

Consider Molarity as concentration

$\therefore ∆T_f=1.86\times \frac{0.6}{4}=0.279\mathrm{K}$