The boiling point of a solution of $0.11 g$ of a substance in $15 g$ of ether was found to be ${0.1}^{o} C$ higher than that of pure ether. The molecular weight of the substance will be: $(K_{b} = 2.16 K k g m o l^{- 1})$ .

148

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158

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168

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178

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Solution

The correct option is B $158$
Mass of solute $W_{2} = 0.11 g$
Mass of solvent $W_{1} = 15 g$
Elevation in boiling point $Δ T_{b} = {0.1}^{o} C$
$K_{b} = 2.16 K k g / m o l$
Molecular mass of solute $M_{2} = \frac{100 \times W_{2} \times K_{b}}{Δ T_{b} \times W_{1}}$
$= \frac{1000 \times 0.11 \times 2.16}{0.1 \times 15}$
$= 158.4 g / m o l$

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The boiling point of a solution of 0.11 g of a substance in 15 g of ether was found to be 0.1oC higher than that of pure ether. The molecular weight of the substance will be: (Kb=2.16Kkgmol−1).

The correct option is B 158Mass of solute W2=0.11gMass of solvent W1=15gElevation in boiling point ΔTb=0.1oCKb=2.16Kkg/molMolecular mass of solute M2=100×W2×KbΔTb×W1 =1000×0.11×2.160.1×15 =158.4g/mol

The boiling point of a solution of $0.11 g$ of a substance in $15 g$ of ether was found to be ${0.1}^{o} C$ higher than that of pure ether. The molecular weight of the substance will be: $(K_{b} = 2.16 K k g m o l^{- 1})$ .