Calculate the boiling point elevation of an aqueous solution containing 0.5 g glucose in 20 g water if K b for water is 0.52 K kg mol − 1 . [Molar mass of glucose = 180 g mol − 1 ]

Question

Calculate the boiling point elevation of an aqueous solution containing 0.5 g glucose in 20 g water if K b ​ for water is 0.52 K kg mol − 1 . [Molar mass of glucose = 180 g mol − 1 ]

TestHub · Accepted Answer

The elevation in boiling point is given by the formula:

$Δ T_{b} = K_{b} \times m$

where $m$ is the molality of the solution.

Molality $m = \frac{W _{B} \times 1000}{M _{B} \times W _{A}}$

Given:
$W_{B} = 0.5$ g (mass of glucose)
$M_{B} = 180$ g mol $^{- 1}$ (molar mass of glucose)
$W_{A} = 20$ g (mass of water)
$K_{b} = 0.52$ K kg mol $^{- 1}$

Substituting the values into the molality formula:

$m = \frac{0.5 \times 1000}{180 \times 20} = \frac{500}{3600} = \frac{5}{36}$ mol kg $^{- 1}$

Now, calculating the boiling point elevation:

$Δ T_{b} = 0.52 \times \frac{5}{36} = \frac{2.6}{36} \approx 0.072$ K

Answer: $0.072 K$

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