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Standard XII

Chemistry

Elevation in Boiling Point

A solution co...

Question

A solution containing 3.3 g of a substance in 125 g of benzene (b.pt. = $80^{o} C$ ) boils at ${80.66}^{o} C$ . If $K_{b}$ for benzene is 3.28 K kg $m o l^{- 1}$ the molecular mass of the substance will be:

130.2

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129.2

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132.2

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131.2

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Solution

The correct option is D 131.2
$Δ T b = K b \times m$
$Δ T b = 0.66 = 3.28 K$ kg $m o l^{- 1}$
$m = \frac{W b}{M b} \times \frac{1}{M a s s o f s o l v e n t (k g)}$
$m = \frac{3.3}{M} \times \frac{1}{0.125}$
$0.66 = 3.28 (\frac{3.3}{M} \times \frac{1}{0.125})$
$M = 131.20 g$

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A solution containing 3.3 g of a substance in 125 g of benzene (b.pt. = 80oC) boils at 80.66oC. If Kb for benzene is 3.28 K kg mol−1 the molecular mass of the substance will be:

The correct option is D 131.2ΔTb=Kb×mΔTb=0.66=3.28K kg mol−1m=WbMb×1Massofsolvent(kg)m=3.3M×10.1250.66=3.28(3.3M×10.125)M=131.20g

A solution containing 3.3 g of a substance in 125 g of benzene (b.pt. = $80^{o} C$ ) boils at ${80.66}^{o} C$ . If $K_{b}$ for benzene is 3.28 K kg $m o l^{- 1}$ the molecular mass of the substance will be:

The correct option is D 131.2
$Δ T b = K b \times m$
$Δ T b = 0.66 = 3.28 K$ kg $m o l^{- 1}$
$m = \frac{W b}{M b} \times \frac{1}{M a s s o f s o l v e n t (k g)}$
$m = \frac{3.3}{M} \times \frac{1}{0.125}$
$0.66 = 3.28 (\frac{3.3}{M} \times \frac{1}{0.125})$
$M = 131.20 g$