1.24 g of (molar mass 124 g ) is dissolved in 1 kg of water to form a solution with boiling point of 100.0156°C, while 25.4 g of (molar mass 250 g ) in 2 kg of water constitutes a solution with a boiling point of 100.0260°C. Which of the following is

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Solution

Q.
1.24 g of $A X_{2}$ (molar mass 124 g ${mol}^{- 1}$ ) is dissolved in 1 kg of water to form a solution with boiling point of 100.0156°C, while 25.4 g of $A Y_{2}$ (molar mass 250 g ${mol}^{- 1}$ ) in 2 kg of water constitutes a solution with a boiling point of 100.0260°C.

$K_{b} (H_{2} O) = {0.52 K kg mol}^{- 1}$

Which of the following is correct [2025]

1 ${AX}_{2}$ and $A Y_{2}$ (both) are completely unionised.

2 ${AX}_{2}$ is fully ionised while $A Y_{2}$ is completely unionised.

3 ${AX}_{2}$ and $A Y_{2}$ (both) are fully ionised.

4 ${AX}_{2}$ is completely unionised while $A Y_{2}$ is fully ionised.

Ans.
(2)

$Moles of A X_{2} (n_{A X_{2}}) = \frac{W_{A X_{2}}}{M_{A X_{2}}} = \frac{1.24}{124} mol = 0.01 mol$

$Molality of A X_{2} solution (m_{A X_{2}}) = \frac{n_{A X_{2}}}{W_{water (in kg)}} = \frac{0.01 mol}{1 kg} = 0.01 m$

$Elevation in boiling point of A X_{2} solution (Δ T_{b} (A X_{2})) = (100.0156 - 100) ° C = i_{A X_{2}} m_{A X_{2}} k_{b}$

$0.0156 = i_{A X_{2}} \times 0.01 \times 0.52$

$i_{A X_{2}} = \frac{0.0156}{0.01 \times 0.52} = 3$

So, $A X_{2}$ is completely ionized.

$Moles of A Y_{2} (n_{A Y_{2}}) = \frac{W_{A Y_{2}}}{M_{A Y_{2}}} = \frac{25.4}{250} mol = 0.1016 mol$

$Molality of A Y_{2} solution (m_{A Y_{2}}) = \frac{n_{A Y_{2}}}{W_{water (in kg)}} = \frac{0.1016 mol}{2 kg} = 0.0508 m$

$Elevation in boiling point of A Y_{2} solution (Δ T_{b} (A Y_{2})) = (100.026 - 100) ° C = i_{A Y_{2}} m_{A Y_{2}} k_{b}$

$0.026 = i_{A Y_{2}} \times 0.0508 \times 0.52$

$i_{A Y_{2}} = \frac{0.026}{0.0508 \times 0.52} \approx 1$

So, $A Y_{2}$ is completely unionized.

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