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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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Similar questions

Q. When

20 g

of naphthoic acid (

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

) is dissolved in

50 g

of benzene

(K_{f}

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

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 -

Q. When 20

g

of naphthoic acid is dissolved in 50

g

of benzene

(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:

QUESTION 2.33

19.5 g of $C H_{2} F C O O H$ is dissolved in 500g of water. The depression in the freezing point of water observed in 1.0 C. Calculate the van't Hoff factor and dissociation constant of fluoroacetic acid.
$K_{f} = 1.86 K k g m o l^{- 1}$

Q. 3.9g of benzoic acid dissolved in 49 g of benzene shows a depression in freezing point of 1.62 K. Calculate the Van't Hoff factor and predict the nature of solute (associated or dissociated).
[Given: Molar mass of benzoic acid =122 g

m o l^{-}, k_{f}

for benzene=4.9 K kg

m o l^{- 1}

]

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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