What is the boiling point of solution of $0.1 \mathrm{~m} \mathrm{KCl}$ ? $\mathrm{K}_{\mathrm{b}}$ of water is $0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \cdot(\alpha=100 \%)$(water boil at $373 \mathrm{~K}$ )

Question

Aicharya · Accepted Answer

$\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} 	imes \mathrm{m} 	imes \mathrm{i}=(0.52)(0.1)(2)=0.104 \mathrm{~K}$ $\Rightarrow \mathrm{T}_{\mathrm{b}}=\mathrm{T}_{\mathrm{b}}^0+\Delta \mathrm{T}_{\mathrm{b}}=373+0.104=373.104 \mathrm{~K}$

What is the boiling point of solution of $0.1 \mathrm{~m} \mathrm{KCl}$ ? $\mathrm{K}_{\mathrm{b}}$ of water…

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