Question

A solution of 0.3 g of benzoic acid in 20 g benzene freeze at 0.317°C below the freezing point of the solvent calculate da degree of association of the acid assuming it to exist in the dimeric state apprant molar mass of the acid kf =5.1 kkg mol-¹.​

Answer 1

2g of benzoic acid (C

6

H

5

COOH) dissolved in 25g of benzene shows a depression in freezing point equal to 1.62K. Molal depression constant for benzene is 4.9 kg m

−1

. What is the percentage association of acid if it forms a dimer in solution?

Hard

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Answer

Correct option is

A

99.2 %

Given,

W

B

=2g,K

f

=4.9KKgmol

−1

W

A

=25g,ΔT

f

=1.62K

Now,

ΔT

f

=K

f

×

M

B

W

B

×

W

A

1000

M

B

=

1.62×25

4.9×2×1000

=241.98g/mol

2C

6

H

5

COOH⇌(C

6

H

5

COOH)

2

If x is the degree of association, (1−x) mole of benzoic acid left undissociated & corresponding x/2 as associated moles of C

6

H

5

COOH at equilibrium.

∴ Total number of moles of particles at eqm,

⇒1−x+

2

x

=1−

2

x

=i

⇒i=

Abnormalmolecularmass

Normalmolecularmass

⇒1−

2

x

=

241.98

122

⇒

2

x

=1−

241.98

122

⇒x=0.992=99.2%

∴ Degree of association of benzene =99.2%

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Enjoy a better ,

W

B

=2g,K

f

=4.9KKgmol

−1

W

A

=25g,ΔT

f

=1.62K

Now,

ΔT

f

=K

f

×

M

B

W

B

×

W

A

1000

M

B

=

1.62×25

4.9×2×1000

=241.98g/mol

2C

6

H

5

COOH⇌(C

6

H

5

COOH)

2

If x is the degree of association, (1−x) mole of benzoic acid left undissociated & cor

I think it's helps to u

Answer 2

The degree of association of the acid assuming it to exist in the dimeric state is 98.8%.

Given:-

Mass of Benzoic Acid (M1) = 0.3g

Mass of Benzene (M2) = 20g

Freezing Point (Tf) = 0.317°C

kf = 5.1kkg mol-¹.

To Find:-

The degree of association of the acid assuming it to exist in the dimeric state.

Solution:-

You can simply calculate the degree of association of the acid assuming it exists in the dimeric state by following these simple steps.

As

Mass of Benzoic Acid (M1) = 0.3g

Mass of Benzene (M2) = 20g

Freezing Point (Tf) = 0.317°C = ∆Tf

kf = 5.1kkg mol-¹.

The molar mass of Benzoic Acid = 122

According to the formula,

$∆Tf = kf \times m$

$∆Tf = kf \times \frac{M1 \times 1000}{M2 \times mm}$

on putting all the values,

$0.317 = 5.1 \times \frac{0.3 \times 1000}{20 \times mm}$

$mm = \frac{5.1 \times 0.3 \times 1000}{0.317 \times 20}$

$mm = \frac{1530}{6.34}$

$mm = 241.32$

Now, as it forms a dimer, i = 1-x/2

$i = \frac{normal \: molar \: mass}{abnormal \: molar \: mass}$

$1 - \frac{x}{2} = \frac{normal \: molar \: mass}{abnormal \: molar \: mass}$

$\frac{x}{2} = 1 - \frac{122}{241.2}$

$\frac{x}{2} = 0.494$

$x = 0.988$

$x = 98.8\%$

Hence, The degree of association of the acid assuming it to exist in the dimeric state is 98.8%.

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