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Depression in Freezing Point

75.2g of C6...

Question

$75.2 g$ of $C_{6} H_{5} O H$ (Phenol) is dissolved is $960 g$ of a solvent of $K_{f} = 14 K k g {k g}^{- 1}$ . If the depression in the freezing point is $7 K$ , then find the percentage of phenol that dimerizes.

75

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80

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30

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40

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Solution

The correct option is B $75$
Please find the answer to your question
$2 C_{6} H_{5} O H ⇌ (C_{6} H_{5} O H)_{2}$
Initial: $1$ $0$
At eqb $1 - α$ $α / 2$
Total number of moles at equilibrium $= 1 - α + α / 2 = 1 - α / 2$
$Δ T_{f} = i K_{f} m$
⇒ $7 = 14 \times 75.2 / 94 \times (1 - α / 2)$ [weight of phenol = 75.2g mol. wt of phenol = 94]
∴ $α = 0.75$
So the percentage of phenol that dimerises $= 75$ %

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Similar questions

75.2 g

K_{f} = 14

7 K

75.2 g C_{6} H_{5} O H

1

K_{f} = 14 K m o l a l i t y^{- 1}

7 K

C_{6} H_{5} O H

K_{f} = 14 K m o l a l^{- 1}

75.2 g

C_{6} H_{5} O H (Phenol)

K_{f} = 14

7 K

molar mass of phenol = 94 {g mol}^{- 1}

= 1 kg

C_{6} H_{5} O H

K_{f} = 14

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