Elevation in boiling point was ${0.52}^{o}$ C when $6$ g of a compound $X$ was dissolved in $100$ g of water. The molecular weight of $X$ is:
[ $K = 5.2 K m o l^{- 1} per 100 g H_{2} O$ ]

120

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60

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180

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342

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Solution

The correct option is B $60$
Elevation is boiling point is given as:

$Δ T_{b} = K_{b} \times m$ $m =$ molality

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

$W_{2} = Mass of solute,$
$W_{1} = Mass of solvent,$
$M_{2} = Molar mass of solute$

$M_{2} = \frac{K_{b} \times W_{2} \times 1000}{W_{1} \times Δ T_{b}}$ as, $K_{b} = 5.2$ per $100 g$ of water

$\Rightarrow M_{2} = \frac{5.2 \times 6 \times 1000}{100 \times 0.52 \times 10}$ $K_{b} = 0.52$ per $k g$ of water

$\Rightarrow M_{2} = 60$ .

Hence, the correct option is $B$

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Elevation in boiling point was 0.52oC when 6g of a compound X was dissolved in 100g of water. The molecular weight of X is: [K=5.2 K mol−1 per 100 g H2O]

Elevation in boiling point was ${0.52}^{o}$ C when $6$ g of a compound $X$ was dissolved in $100$ g of water. The molecular weight of $X$ is:
[ $K = 5.2 K m o l^{- 1} per 100 g H_{2} O$ ]