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Standard XII

Chemistry

Van't Hoff Factor

When 20 g of ...

Question

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

0.5

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Solution

The correct option is A 0.5
$Δ T f = i \times K f \times m$

or $2 = i \times 1.72 \times \frac{20}{172} \times \frac{1}{50} \times 1000$
or i =0.5

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Q. When 20 g of naphthoic acid

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is dissolved in 50 g of benzene

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

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

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

The correct option is A 0.5ΔTf=i×Kf×m or 2=i×1.72×20172×150×1000 or i =0.5

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

The correct option is A 0.5
$Δ T f = i \times K f \times m$

or $2 = i \times 1.72 \times \frac{20}{172} \times \frac{1}{50} \times 1000$
or i =0.5