Which of the following aqueous solutions should have the least elevation in boiling point?

1.0 M K O H

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1.0 M (N H_{4})_{2} S O_{4}

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1.0 M K_{2} C O_{3}

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1.0 M K_{2} S O_{4}

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Solution

The correct option is A $1.0 M K O H$
From elevation in boiling point,
$Δ T_{b} = i \times K_{b} \times m$
where,
$△ T_{b}$ is the elevation in boiling point.
$K_{b}$ is molal elevation constant.
m is molality of solution.
i is van't Hoff factor

From the equation we can say lower the value of $i \times m$ lower will be the boiling point elevation.
Considering all the compounds exhibit 100% dissociation,

$\begin{matrix} K O H \Rightarrow & i \times m = 2 \times 1 = 1 (N H_{4})_{2} S O_{4} \Rightarrow & i \times m = 3 \times 1 = 3 K_{2} C O_{3} \Rightarrow & i \times m = 3 \times 1 = 3 K_{2} S O_{4} \Rightarrow & i \times m = 3 \times 1 = 3 \end{matrix}$
Since, $i \times m$ is lower for $K O H$ , the boiling point will be lower for $1 M K O H$

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