Question

The $0.1M$ of $KI$ and $0.2M$ of $AgN{O_3}$ are mixed in $3:1$ volume ratio. The depression of freezing point of the resulting solution will be $\left[ {{K_f}\left( {{H_2}O} \right) = 1.86K - Kg{{\left( {mol} \right)}^{ - 1}}} \right]$
A.$3.72K$
B.$1.86K$
C.$0.93K$
D.$0.279K$

Answer 1

The change in freezing point is one of the colligative properties and can be determined from the Van’t Hoff factor, molality and molal freezing point constant. Molal freezing point is constant and the molality is determined from the number of moles and volume of solution.

Complete answer:
The change in freezing point is also known as depression in freezing point.
Given that $0.1M$ of $KI$ and $0.2M$ of $AgN{O_3}$ are mixed in $3:1$ volume ratio. Thus, the number of moles will be $0.3mol$ of $KI$ and $0.2mol$ of $AgN{O_3}$. Out of these $0.3mol$ of $KI$ and $0.2mol$ of $AgN{O_3}$, $0.2mol$ of $KI$ and $0.2mol$ of $AgN{O_3}$ are reacted with each other to form potassium nitrate and silver iodide.
$AgN{O_3} + KI \to AgI + KN{O_3}$

$0.3mol$	$0.2mol$	-	-
$0.1mol$	-	$0.2mol$	$0.2mol$

Thus, the number of moles of $KN{O_3}$ and $KI$ will be $0.2 \times 2 + 0.1 \times 2 = 0.6$ The volume of solution given is $4$, the molality will be $\dfrac{{0.6}}{4}$ and Van’t Hoff factor is two for two salts.
The total molality will be $\dfrac{{0.6}}{4}$
The molal depression freezing point is already given as $1.86K - Kg{\left( {mol} \right)^{ - 1}}$
$\Delta {T_f} = {K_f} \times m$
Substitute these values in the above formula:
$\Delta {T_f} = 1.86 \times \dfrac{{0.6}}{4} = 0.279K$
When $0.1M$ of $KI$ and $0.2M$ of $AgN{O_3}$ are mixed in $3:1$ volume ratio. The depression of freezing point of the resulting solution will be $0.279K$
Hence, option (D) is the correct answer.

Note:
While considering the molality, the number of moles of each solute must be calculated. As the number of moles were calculated, these should multiply with the number of ions given the value of the number of moles of mixture and total volume of mixture is $4$. Thus, final molality will be the number of moles of mixture divided by total volume of mixture.