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Abnormal Colligative Properties

The boiling p...

Question

The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. $(K_{b} f o r b e n z e n e i s 2.53 K k g m o l^{- 1})$

$58.0 g m o l^{- 1}$

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$68.0 g m o l^{- 1}$

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$78.0 g m o l^{- 1}$

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$48.0 g m o l^{- 1}$

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Solution

The correct option is A
$58.0 g m o l^{- 1}$

Given values are:
$T_{b e n z e n e}^{\circ} = 353.00 K; K_{b} = 2.53 K k g m o l^{- 1}$
$T_{b (s o l u t i o n)} = 354.00 k$
$W_{s o l u t e} = 1.80 g$
$W_{s o l v e n t} = 90 g$
The elevation in boiling point, $Δ T_{b} = T_{b (s o l u t i o n)} - T_{b}^{\circ} (s o l v e n t)$
$= 354.11 - 353.23$
$= 0.88 K$
Molar mass of solute is given as
$M w_{s o l u t e} = \frac{K_{b} \times 1000 \times W_{s o l v e n t}}{Δ T_{b} \times W_{s o l v e n t}}$
$M w_{s o l u t e} = \frac{2.53 \times 1000 \times 1.80}{0.88 \times 90} = 58.0 g m o l^{- 1}$
Hence, the molar mass of solute is $58.0 g m o l^{- 1}$

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The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. (Kb for benzene is 2.53Kkgmol−1)

The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. $(K_{b} f o r b e n z e n e i s 2.53 K k g m o l^{- 1})$