Pure benzene freezes at 5-3- C- A solution of 0-223 g of phenylacetic acid C 6 H 5 CH 2 COOH in 4-4 g of benzene K f -5-12 Kkg mol -1freezes at 4-47- C- From the observation- one can conclude thatA- Phenylacetic acid exists as it is in benzeneB- Phenylacetic acid undergoes partial ionisation in benzeneC- Phenylacetic acid undergoes complete ionisation in benzeneD- Phenylacetic acid dimerises in benzene

The correct option is D Phenylacetic acid dimerises in benzene nbsp;Δ T_f=T_0 T_f=5.3 4.47=0.83 nbsp; Molality of solvent, m = 0.223/136×1000/4.4=0.373 ...

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Byju's Answer

Standard XII

Chemistry

Depression in Freezing Point

Pure benzene ...

Question

Pure benzene freezes at ${5.3}^{\circ} C$ . A solution of $0.223 g$ of phenylacetic acid $(C_{6} H_{5} C H_{2} C O O H)$ in $4.4 g$ of benzene $(K_{f} = 5.12 {Kkg mol}^{- 1})$ freezes at ${4.47}^{\circ} C$ . From the observation, one can conclude that

Phenylacetic acid exists as it is in benzene

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Phenylacetic acid undergoes partial ionisation in benzene

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Phenylacetic acid undergoes complete ionisation in benzene

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Phenylacetic acid dimerises in benzene

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Solution

The correct option is D Phenylacetic acid dimerises in benzene
$Δ T_{f} = T_{0} - T_{f} = 5.3 - 4.47 = 0.83$
Molality of solvent, $m = \frac{0.223}{136} \times \frac{1000}{4.4} = 0.373$ (Molecular weight of phenyl acetic acid $= 136 g m o l^{- 1}$
$K_{f} = 5.12$
$Δ T_{f} = i . K_{f} m$
$i = (Δ T_{f}) / (K_{f} m) = \frac{0.83}{5.12 \times 0.373} \approx 0.43$

It is nearly equal to 0.5 which means that the phenyl acetic acid undergoes dimerization in benzene.

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Similar questions

Q. Pure benzene freezes at

{5.3}^{\circ} C

. A solution of

0.223 g

of phenylacetic acid

(C_{6} H_{5} C H_{2} C O O H)

4.4 g

of benzene

(K_{f} = 5.12 {Kkg mol}^{- 1})

freezes at

{4.47}^{\circ} C

. From the observation, one can conclude that

Q. For the given reaction :

2-phenylacetic acid +NaOH C a O, △ ⟶ Product

The product is:

Q. Pure benzene freezes at

{5.3}^{o} C

. A solution of 0.223 g of phenyl acetic acid (

C_{6} H_{5} C H_{2} C O O H

) in 4.4 g of benzene (

K_{f}

5.12 K kg

m o l^{- 1}

) freezes at

{4.47}^{o} C

. From the observation one can conclude that:

How will you prepare the following compounds from benzene? You may use any inorganic reagent and any organic reagent having not more than one carbon atom

(i) Methyl benzoate (ii) m-Nitrobenzoic acid

(iii) p-Nitrobenzoic acid (iv) Phenylacetic acid

(v) p-Nitrobenzaldehyde.

Q. How will you prepare the following compounds from benzene? You may use any inorganic reagent and any organic reagent having not more than one carbon atom:
(i) Methyl benzoate (ii) m-Nitrobenzoic acid
(iii) p-Nitrobenzoic acid (iv) Phenylacetic acid
(v) p-Nitrobenzaldehyde.

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Pure benzene freezes at 5.3∘C. A solution of 0.223 g of phenylacetic acid (C6H5CH2COOH) in 4.4 g of benzene (Kf=5.12 Kkg mol−1) freezes at 4.47∘C. From the observation, one can conclude that

Pure benzene freezes at ${5.3}^{\circ} C$ . A solution of $0.223 g$ of phenylacetic acid $(C_{6} H_{5} C H_{2} C O O H)$ in $4.4 g$ of benzene $(K_{f} = 5.12 {Kkg mol}^{- 1})$ freezes at ${4.47}^{\circ} C$ . From the observation, one can conclude that