When 20 g of naphthoic acid (C11H8O2) is dissolved in 50 g of benzene (Kf = 1.72 K kg mol-1), a freezing point depression of 2 K is observed. The van-t Hoff factor (i) is

Question

When 20 g of naphthoic acid (C11H8O2) is dissolved in 50 g of benzene (Kf = 1.72 K kg mol-1), a freezing point depression of 2 K is observed. The van-t Hoff factor (i) is (A) 0.5 (B) 1 (C) 2 (D) 3

Tardigrade Admin · Accepted Answer

Molality=((20/172)) × (1000/50) =2.325 m ⇒ -Δ Tf =2 = iKf.m ⇒ i=(2/1.72 × 2.325)=0.5

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 $2 K$ is observed. The van-t Hoff factor (i) is

0.5

1

2

3

Molality $= (\frac{20}{172}) \times \frac{1000}{50} = 2.325 m$
$\Rightarrow - Δ T_{f} = 2 = i K_{f} . m$
$\Rightarrow i = \frac{2}{1.72 \times 2.325} = 0.5$

Q. When 20g of naphthoic acid (C11​H8​O2​) is dissolved in 50g of benzene (Kf​=1.72Kkgmol−1), a freezing point depression of 2K is observed. The van-t Hoff factor (i) is

0.5

1

2

3

Molality=(17220​)×501000​=2.325m ⇒−ΔTf​=2=iKf​.m ⇒i=1.72×2.3252​=0.5

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 $2 K$ is observed. The van-t Hoff factor (i) is

Molality $= (\frac{20}{172}) \times \frac{1000}{50} = 2.325 m$
$\Rightarrow - Δ T_{f} = 2 = i K_{f} . m$
$\Rightarrow i = \frac{2}{1.72 \times 2.325} = 0.5$