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Standard XII

Chemistry

Abnormal Colligative Properties

When 20 g of ...

Question

When 20 g of naphthoic acid ( $C_{11} H_{8} O_{2}$ ) is dissolved in 50 g of benzene ( $K_{f} = 1.72 K K g m o l^{- 1}$ ), a freezing point depression of 2K is observed. The van’ t Hoff factor (i) is

0.5

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Solution

The correct option is A 0.5
$M o l a l i t y = (\frac{20}{172}) \times \frac{1000}{50} = 2.325 m$
$\Rightarrow - Δ T_{f} = 2 = i K_{f} . m$
$\Rightarrow i = \frac{2}{1.72 \times 2.325} = 0.5$

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Similar questions

Q. When 20 g of naphthoic acid (

C_{11} H_{8} O_{2}

) is dissolved in 50 g of benzene (

K_{f} = 1.72 K K g m o l^{- 1}

), a freezing point depression of 2K is observed. The van’ t Hoff factor (i) is

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C_{1} 1 H_{8} O_{2}

) is dissolved in

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freezing point depression of

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is observed. The van't Hoff factor

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When 20 g of naphthoic acid (C11H8O2) is dissolved in 50g of benzene (Kf = 1.72 K kg mol-1), a freezing point depression of 2K is observed. The van't Hoff factor (i) is -

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(K_{f} = 1.72 K k g m o 1^{- 1})

, a freezing point depression of 2

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is observed. The van't Hoff factor (i) is:

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When 20 g of naphthoic acid (C11H8O2) is dissolved in 50 g of benzene ( Kf=1.72KKgmol−1), a freezing point depression of 2K is observed. The van’ t Hoff factor (i) is

The correct option is A 0.5Molality=(20172)×100050=2.325m ⇒ −ΔTf=2=iKf.m ⇒ i=21.72×2.325=0.5

When 20 g of naphthoic acid ( $C_{11} H_{8} O_{2}$ ) is dissolved in 50 g of benzene ( $K_{f} = 1.72 K K g m o l^{- 1}$ ), a freezing point depression of 2K is observed. The van’ t Hoff factor (i) is

The correct option is A 0.5
$M o l a l i t y = (\frac{20}{172}) \times \frac{1000}{50} = 2.325 m$
$\Rightarrow - Δ T_{f} = 2 = i K_{f} . m$
$\Rightarrow i = \frac{2}{1.72 \times 2.325} = 0.5$