JEE MainChemistryToughPhysical Chemistry

To 500 cm3 of water, 3.0 × 10-3 kg of acetic acid is added. If
23 of acetic acid is dissociated, what will be the depression in
freezing point? kf and density of water are 1.86 K kg mole-1 and 0.997
g cm-3 respectively
500 cm3 ty esa 3.0 × 10-3 kg,flfVd vEy Mkyk A;fn 23,flfVd
vEy rks fgekad esa fdruk voueu gksxk kf o ty dk
?kuRo Øe'k 1.86 K kg mole-1 o 0.997 g cm-3 gaS

Show answer and solution

Weight of water

= 500 \times 0.997 = 498.5g

No. of moles of acetic acid

= \frac{\ Wt.\ of\ {CH}_{3}COOH\ (ingm)\ }{\ Mol.wt.of\ {CH}_{3}COOH} = \frac{3 \times 10^{- 3} \times 10^{3}}{60} = 0.05

Since 498.5 g of water has 0.05 moles of CH3COOH
1000 g of water has

= \frac{0.05 \times 1000}{498.5} = 0.1

Determination of van`t Hoff factor, i

{CH}_{3}COOH \rightarrow {CH}_{3}{COO}^{-} + H^{+}

No. of moles at start 1 0 0
No. of moles at equb. 1-0.23 0.23 0.23

\Delta T_{r} = (1 - 0.23 + 0.23 + 0.23) \times 1.86 \times 0.1 = 0.228K

Hindi ty dk Hkkj

= 500 \times 0.997 = 498.5g

,lhfVd vEy ks

= \frac{\ Wt.\ of\ {CH}_{3}COOH\ (ingm)\ }{\ Mol.wt.of\ {CH}_{3}COOH} = \frac{3 \times 10^{- 3} \times 10^{3}}{60} = 0.05

498.5 g ty esa CH3COOH ds 0.05 eksy gksxasA
1000 g ty

= \frac{0.05 \times 1000}{498.5} = 0.1

okWUVgkWQ xq.kkad fy, eku.k

{CH}_{3}COOH \rightarrow {CH}_{3}{COO}^{-} + H^{+}

çkjEHk ksa 1 0 0
ks 1-0.23 0.23 0.23

\Delta T_{r} = (1 - 0.23 + 0.23 + 0.23) \times 1.86 \times 0.1 = 0.228K

Source: Grass-G / CH-26 Chemistry Paper 1 15 Jan.docx

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