What is the boiling point of an aqueous solution containing $117$ g NaCl in 1000 g H $_{2}$ O?
[K $_{b} =$ 0.52 kg/ mol]

98.96

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99.48

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100.52

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101.04

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102.08

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Solution

The correct option is E $102.08$
The molar mass of $N a C l$ is 58.4 g/mol.

The number of moles of $N a C l = \frac{117 g}{58.4 g / m o l} = 2 m o l$

Mass of water $= 1000 g = 1 k g$

The molality of $N a C l$ $= \frac{2 m o l}{1 k g} = 2 m$

$N a C l ⇌ N a^{+} + C l^{-}$

The vant Hoff factor $i = 2$

The elevation in the boiling point

$Δ T_{b} = i K_{b} m = 2 \times 0.52 \times 2 = 2.08^{o} C$

The boiling pint of pure water is 100 deg C.

The boiling point of aqueous solution $= 100 + 2.08 = 102.08^{o} C$ .

Hence, the correct option is $E$

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What is the boiling point of an aqueous solution containing 117 g NaCl in 1000 g H2O? [Kb= 0.52 kg/ mol]

What is the boiling point of an aqueous solution containing $117$ g NaCl in 1000 g H $_{2}$ O?
[K $_{b} =$ 0.52 kg/ mol]