Question

$30 mL$ of $0.1 M KI ( aq )$ and $10 mL$ of $0.2 M AgNO _{3}$ are mixed. The solution is then filtered out. Assuming no change in total volume, depression in freezing point of the resulting solution will be $y \times 10^{-2}{ }^{\circ} C$. Value of $y$ : (Given: $K _{ f }$ for $H _{2} O =1.86 K \quad kg mol m ^{-1}$, assume molarity = molality)

Answer 1

$\therefore $ [KI] in solution $= \frac{1}{4 0}$ and [KNO₃] $= \frac{2}{4 0}$ &
(Assuming molarity = molality for dilute soluation)
$\therefore \, \, \text{Δ} \left(\text{T}\right)_{\text{f}} \, = \, \text{Δ} \left(\text{T}\right)_{\text{f}} \, \, \left(\right. \text{b} \text{y} \, \text{K} \text{I} \left.\right) \, \, + \, \, \text{Δ} \left(\text{T}\right)_{\text{f}} \, \, \left(\right. \text{b} \text{y} \, \text{K} \text{N} \left(\text{O}\right)_{3} \left.\right)$
$= \text{molality} \times \text{1.86} \times \text{i}_{\text{KI}} + \text{molality} \times \text{1.86} \times \text{i}_{\text{KNO}_{3}}$
$= \left(\right. \frac{1}{40} \times 1.86 \times 2 \left.\right) + \left(\right. \frac{2}{40} \times 1.86 \times 2 \left.\right)$
$= 0 \text{.} 0 9 3 + 0 \text{.} 1 8 6 = 0 \text{.} 2 7 9 = 0.28 ^\circ \text{C}$