An aqueous solution freezes at $- {2.55}^{\circ} C$ . What will be the boiling point of solution?
$(K_{b} (H_{2} O) = 0.52 K / m; K_{f} (H_{2} O) = 1.86 k / m)$

{107.0}^{\circ} C

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{100.6}^{\circ} C

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{100.1}^{\circ} C

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100.7 \circ C

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Solution

The correct option is D $100.7 \circ C$
$Δ T_{f} = K_{f}; m = m o l a l i t y 2.55 = 1.86 \times m$
m = 1.37
$Δ T_{b} = K_{b} . m = 0.52 \times 1.37$
= 0.712
$T_{b} = T_{b}^{\circ} + Δ T_{b}$
= 100 + 0.712 = ${100.7}^{\circ} C$

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Similar questions

An aqueous solution freezes at $- {2.55}^{\circ} C$ . What is its boiling point?

$(K_{b} H_{2} O = 0.52 K / m; K_{f} H_{2} O = 1.86 K / m)$

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