Question

A solution containing

0.5126 g

of naphthalene

(

molar mass =

128.17 g {mol}^{- 1}

)

in

50.00 g

of

{CCl}_{4}

gives a boiling point elevation of

0.402 K

. While a solution of

0.6216 g

of unknown solute in the same mass of the solvent gives a boiling point elevation of

0.647 K

. Find the molar mass of the unknown solute.

Accepted Answer

Mass of naphthalene, $W_{2} = 0.5126 g = 0.5126 \times 10^{- 3} kg$

Molar mass of naphthalene, $M_{2} = 128.17 g {mol}^{- 1}$

Mass of carbon tetrachloride, $W_{1} = 50 g = 50 \times 10^{- 3} kg$

$∆ T_{b 1} = 0.402 K$

Calculation of $K_{b}$ for ${CCl}_{4}$ -

$K_{b} = \frac{∆ T_{b 1} \times W_{1} \times M_{2}}{1000 \times W_{2}} = \frac{0.402 \times 50 \times 10^{- 3} \times 128.17}{1000 \times 0.5126 \times 10^{- 3}} = 5.03 K kg {mol}^{- 1}$

Molal boiling point elevation constant, $K_{b} = 5.03 K kg {mol}^{- 1}$

$∆ T_{b 2} = 0.647 K$

Mass of unknown solute, $W_{2}' = 0.6216 g = 0.6216 \times 10^{- 3} kg$

$K_{b} = \frac{∆ T_{b 2} \times W_{1} \times M_{2}'}{1000 \times W_{2}'}$

Therefore, $M_{2}' = \frac{K_{b} \times W_{2}' \times 1000}{∆ T_{b 2} \times W_{1}} = \frac{5.03 \times 0.6216 \times 10^{- 3} \times 1000}{0.647 \times 50 \times 10^{- 3}}$

$= 96.65 g {mol}^{- 1}$

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