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Standard XII

Chemistry

Elevation in Boiling Point

The boiling p...

Question

The boiling point of a solution of 0.105 gm of a substance in 15.84 gram of ether was found to be 100°C higher than that of pure ether. What is the molecular weight of the substance [Molecular elevation constant of ether per 100 g = 21.6]

144.50

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143.18

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140.28

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146.66

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Solution

The correct option is
C

143.18
Mass of substance $W_{b}$ = 0.1050 gram
Molar mass of substance = m
Mass of ether = (solvent) $W_{A}$ = 15.84 gm = 0.1584 kg
$m = \frac{K_{b} \times W_{b} \times 1000}{Δ T_{b} \times W_{A}}$
$K_{b} = 21.6 p e r 100 g m o r 2.16 p e r 1000 g m$
$m o l a l i t y = \frac{2.16 \times 0.105 \times 1000}{0.100 \times 15.84} = 143.18 g m$

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The boiling point of a solution of 0.105 gm of a substance in 15.84 gram of ether was found to be 100°C higher than that of pure ether. What is the molecular weight of the substance [Molecular elevation constant of ether per 100 g = 21.6]

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The boiling point of a solution of 0.105 gm of a substance in 15.84 gram of ether was found to be 100°C higher than that of pure ether. What is the molecular weight of the substance [Molecular elevation constant of ether per 100 g = 21.6]

The correct option is C 143.18Mass of substance Wb = 0.1050 gramMolar mass of substance = mMass of ether = (solvent) WA = 15.84 gm = 0.1584 kgm=Kb×Wb×1000ΔTb×WAKb=21.6per100gmor2.16per1000gm molality=2.16×0.105×10000.100×15.84=143.18gm