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Elevation in Boiling Point

An aqueous so...

Question

An aqueous solution containing $288 g$ of a non-volatile compound having the stoichimetric composition $C_{x} H_{2 x} O_{x}$ in $90 g$ water boils at ${101.24}^{\circ} C$ at one atmospheric pressure. What is the molecular formula of the compound?
$K_{b} (H_{2} O) = 0.512 K m o l^{- 1} k g, T_{b} (H_{2} O) = 100^{\circ} C$

A

C_{44} H_{88} O_{44}

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B

C_{12} H_{24} O_{12}

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C

C_{22} H_{44} O_{22}

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D

C_{10} H_{20} O_{10}

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Solution

The correct option is A $C_{44} H_{88} O_{44}$
Boiling point elevation of a solution is given by,
$△ T_{b} = K_{b} \times m$
where,
$△ T_{b}$ is the elevation in boiling point.
$K_{b}$ is molal elevation constant.
m is molality of solution.

Given:
$Δ T = {1.24}^{o} C$
$K_{b} (H_{2} O) = 0.512 K m o l^{- 1} k g$
Mass of solute $= 288 g$
Mass of solvent $= 90 g$
Molar mass of solute is 'M'

To calculate the molecular mass of
$C_{x} H_{2 x} O_{x}$

$△ T_{f} = \frac{K_{f} \times \frac{w_{solute}}{M} \times 1000}{W_{solvent}}$

$1.24 = (0.512) \times \frac{\frac{288}{M}}{90} \times 1000$

$M = \frac{(288) (1000) (0.512)}{(90) (1.24)}$

$M = 1321.2 g$

To calculate the molecular formula of
$C_{x} H_{2 x} O_{x}$
$12 x + 2 x + 16 x = 1321.1$

$x = 44$
$∴$
$C_{44} H_{88} O_{44}$

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Elevation in Boiling Point

Standard XII Chemistry

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