1.2 mL of acetic acid is dissolved in water to make 2.0 L of solution. The depression in freezing point observed for this strength of acid is 0.0198° C. The percentage of dissociation of the acid is (Nearest integer) [Given: Density of acetic acid is 1.02 g mL -1 Molar mass of acetic acid is 60 g mol -1 . K f ( H 2 O )=1.85 K kg mol -1]

Question

1.2 mL of acetic acid is dissolved in water to make 2.0 L of solution. The depression in freezing point observed for this strength of acid is 0.0198° C. The percentage of dissociation of the acid is (Nearest integer) [Given: Density of acetic acid is 1.02 g mL -1 Molar mass of acetic acid is 60 g mol -1 . K f ( H 2 O )=1.85 K kg mol -1] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

M = d × V =1.02 × 1.2=1.224 gm Moles of acetic acid =0.0204 moles in 2 L So molality =0.0102 mol / kg Now Δ T f = i × K f × M i=1+α for acetic acid 0.0198=(1+α) × 1.85 × 0.0102 α=0.04928 ≅ 5 %

M=d×V=1.02×1.2=1.224gm Moles of acetic acid =0.0204 moles in 2L So molality =0.0102mol/kg Now ΔTf​=i×Kf​×M i=1+α for acetic acid 0.0198=(1+α)×1.85×0.0102 α=0.04928 ≅5%

$M = d \times V = 1.02 \times 1.2 = 1.224 g m$
Moles of acetic acid $= 0.0204$ moles in $2 L$
So molality $= 0.0102 m o l / k g$
Now $Δ T_{f} = i \times K_{f} \times M$
$i = 1 + α$ for acetic acid
$0.0198 = (1 + α) \times 1.85 \times 0.0102$
$α = 0.04928$
$≅ 5%$