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Standard XII

Chemistry

Van't Hoff Factor

The freezing ...

Question

The freezing point depression constant for water is ${1.86}^{o} C$ $m^{- 1}$ . If $5$ g $N a_{2} S O_{4}$ is dissolved in 45 g $H_{2} O$ , the freezing point is changed by $- {3.82}^{o} C$ .
Calculate the van't Hoff factor for $N a_{2} S O_{4}$ .

0.381

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2.05

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2.63

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3.11

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Solution

The correct option is C $2.63$
Given: $K_{f} = {1.86}^{o} C m^{- 1}$ , $w_{B} = 5 g, w_{A} = 45 g, M_{B} = 142 g / m o l$
$△ T = i \times K_{f} \times \frac{w_{B} \times 1000}{M_{B} \times w_{A}}$

$3.82 = i \times 1.86 \times \frac{5 \times 1000}{142 \times 45}$

$i = 2.63$

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Similar questions

Q. The freezing point depression constant for water ids

{1.86}^{o} c m^{- 1}

. If

5.0

g of

N a_{2} S O_{4}

is dissolved in

45.0

H_{2} O

, the freezing point is change by

- {3.82}^{o} C

. Calculate the van't Hoff factor for

N a_{2} S O_{4}

Q. If

5 g

N a_{2} S O_{4}

is dissolved in

45 g

H_{2} O

, the freezing point is changed by

{3.72}^{\circ} C

. Calculate the van't Hoff factor for

N a_{2} S O_{4}

.
Given: Freezing point depression constant for water is

{1.86}^{\circ} C k g m o l^{- 1}

Q. The freezing point depression constant for water is

- {1.86}^{0} C m^{- 1} .

If 5.00 g

N a_{2} S O_{4}

is dissolved in 45.0 g

H_{2} O,

the freezing point is changed by

{3.82}^{0} C .

Calculate the vant Hoff factor for

N a_{2} S O_{4} .

Q. Calculate the freezing point depression expected for

0.0711

m aqueous solution of

N a_{2} S O_{4}

. If this solution actually freezes at

- {0.320}^{o}

C, what would be the value of van't Hoff factor? (

K_{f}

for water is

1.86

K kg

m o l^{- 1}

Q. What mass of

N a C

l must be dissolved in 65.0 g of water to lower the freezing of water by

{7.5}^{o} C

? The freezing point depression constant (

K_{f}

) for water 1.860 C/m. Assume Van't Hoff factor for

N a C l

is 1.87. (Molar mass of

N a C l =

58. 5)

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The freezing point depression constant for water is 1.86oC m−1. If 5 g Na2SO4 is dissolved in 45 g H2O, the freezing point is changed by −3.82oC. Calculate the van't Hoff factor for Na2SO4.

The correct option is C 2.63Given: Kf=1.86oCm−1, wB=5g,wA=45g,MB=142g/mol△T=i×Kf×wB×1000MB×wA3.82=i×1.86×5×1000142×45i=2.63

The freezing point depression constant for water is ${1.86}^{o} C$ $m^{- 1}$ . If $5$ g $N a_{2} S O_{4}$ is dissolved in 45 g $H_{2} O$ , the freezing point is changed by $- {3.82}^{o} C$ .
Calculate the van't Hoff factor for $N a_{2} S O_{4}$ .

The correct option is C $2.63$
Given: $K_{f} = {1.86}^{o} C m^{- 1}$ , $w_{B} = 5 g, w_{A} = 45 g, M_{B} = 142 g / m o l$
$△ T = i \times K_{f} \times \frac{w_{B} \times 1000}{M_{B} \times w_{A}}$

$3.82 = i \times 1.86 \times \frac{5 \times 1000}{142 \times 45}$

$i = 2.63$