Elevation in boiling point was ${0.52}^{o} C$ when 6 g of compound X was dissolved in 100 g of water. Molecular mass of the compound M in $g m o l^{- 1}$ is: ( $K_{b}$ for water is $0.52 K m^{- 1}$ .)

120

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600

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180

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Solution

The correct option is B 60
We have
$Δ T_{b} = K_{b} m$

$0.52 = 0.52 \times \frac{m a s s o f s o l u t e}{m o l a r m a s s o f s o l u t e \times m a s s o f s o l v e n t (i n g)} \times 1000$

$0.52 = 0.52 \times \frac{6}{M \times 100} \times 1000$

$M = 60 g m o l^{- 1}$

Hence, option (B) is correct.

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Elevation in boiling point was 0.52oC when 6 g of compound X was dissolved in 100 g of water. Molecular mass of the compound M in gmol−1 is: (Kb for water is 0.52Km−1.)

The correct option is B 60We haveΔTb=Kbm0.52=0.52×mass of solutemolar mass of solute ×mass of solvent (in g)×10000.52=0.52×6M×100×1000M=60 gmol−1Hence, option (B) is correct.

Elevation in boiling point was ${0.52}^{o} C$ when 6 g of compound X was dissolved in 100 g of water. Molecular mass of the compound M in $g m o l^{- 1}$ is: ( $K_{b}$ for water is $0.52 K m^{- 1}$ .)