Considering acetic acid dissociates in water, its dissociation constant is 6.25 x 10^-5.

Question

Considering acetic acid dissociates in water, its dissociation constant is \(6.25 \times 10 ^{-5}\). If 5 mL of acetic acid is dissolved in 1 litre water, the solution will freeze at \(-x \times 10 ^{-2} \,{^\circ C}\), provided pure water freezes at \(0 ^\circ C\).

x = ______. (Nearest integer)

Given : \((K_f )_{water}\) = 1.86 K kg \(mol ^{-1}.\)

density of acetic acid is \(1.2\,g \, mol ^{-1}\)

molar mass of water = \(18\,g \, mol ^{-1}.\)

molar mass of acetic acid \(=60 \,g\, mol ^{-1}.\)

density of water = \(1\,g \, cm ^{-3}\)

Acetic acid dissociates as

\(CH_3 COOH \rightleftharpoons CH_3 COO ^\odot + H ^\oplus\)

Answer 1

Correct answer is : 19

Mass of acetic acid = 1.2 x 5 = 6 g

Mole of acetic acid = \(\frac {6}{60}\) = 0.1 g

Mass of water = 1000 x 1 = 1000 g

\(\text { Conc }^{\mathrm{n}} \text { of Sol } \mathrm{l}^{\mathrm{n}}=\frac{0.1}{1}=0.1 \mathrm{M} \)

\(\mathrm{k}_{\mathrm{a}}=\mathrm{c} \alpha^2\)

\(\alpha=\left(\frac{6.25 \times 10^{-5}}{10^{-1}}\right)^{1 / 2} \)

\(=\left(625 \times 10^{-6}\right)^{1 / 2}=25 \times 10^{-3}=0.025\)

\(\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{ik}_{\mathrm{f}} \mathrm{m}\)

\(=(1+0.025)(1.86) \times 0.1=0.190\)

\(0^{\circ} \mathrm{C}-\mathrm{T}_{\mathrm{f}}=0.190\)

\(\mathrm{~T}_{\mathrm{f}}=-0.190^{\circ} \mathrm{C}=-19 \times 10^{-2}{ ^\circ }\mathrm{C} \)

\(\mathrm{x}=19\)