The freezing point of a solution containing 0-2 g of acetic acid in 20 g of benzene is lowered by 0-45-C- Calculate the degree of association - for acetic acid-Given -KfBenzene-5-12 K kg mol-1

Weight of acetic acid(w)=0.2 g Weight of benzene(W)=20 g=201000 kg Δ T_f=0.45^∘ C Let m_obs be nbsp;observed molecular mass of acetic acid Δ T_f= K_f× m m ...

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Byju's Answer

Standard XII

Chemistry

Van't Hoff Factor

The freezing ...

Question

The freezing point of a solution containing $0.2 g$ of acetic acid in $20 g$ of benzene is lowered by ${0.45}^{\circ} C .$ . Calculate the degree of association $(β)$ for acetic acid.
Given :
$(K_{f})_{Benzene} = 5.12 K k g m o l^{- 1}$

0.53

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0.81

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0.25

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0.94

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Solution

The correct option is D $0.94$
$Weight of acetic acid (w) = 0.2 g Weight of benzene (W) = 20 g = \frac{20}{1000} k g Δ T_{f} = {0.45}^{\circ} C$

$Let m_{o b s} be observed molecular mass of acetic acid$

$Δ T_{f} = K_{f} \times m$

$m is molality$

$m_{o b s} = \frac{1000 \times K_{f} \times w}{W \times Δ T_{f}} = \frac{1000 \times 5.12 \times 0.2}{20 \times 0.45} = 113.78 g m o l^{- 1}$

$Theoretical molecular mass of acetic acid = 60 g m o l^{- 1}$

$van't Hoff factor = \frac{Theoretical molecular mass}{Observed molecular mass} i = \frac{60}{113.78} \approx 0.53$

$2 C H_{3} C O O H ⇌ (C H_{3} C O O H)_{2} \begin{matrix} Before association & C & 0 After association & (C - C β) & \frac{C β}{2} \end{matrix}$

$i = \frac{C - C β + \frac{C β}{2}}{C} 0.53 = 1 - \frac{β}{2}$
$β = 0.94$

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The freezing point of a solution containing 0.2 g of acetic acid in 20 g of benzene is lowered by 0.45∘C.. Calculate the degree of association (β) for acetic acid. Given : (Kf)Benzene=5.12 K kg mol−1

The freezing point of a solution containing $0.2 g$ of acetic acid in $20 g$ of benzene is lowered by ${0.45}^{\circ} C .$ . Calculate the degree of association $(β)$ for acetic acid.
Given :
$(K_{f})_{Benzene} = 5.12 K k g m o l^{- 1}$