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 at the freezing point? \(K_f\) and density of water are 1.86 K kg−1 mol−1 and 0.997 g cm−3 respectively.
Answer & explanation
Correct answer: option 3
The correct answer is option 3. 0.229.
Mass of solute \(\text{ = 3.0 × }10^{−3} = 3.0 g\)
Molecular mass of the solute (Acetic acid) \(\text{= 60}\)
Mass of the solvent \(\text{ 500 × 0.997 = 498.5 g}\)
We know that,
Degree of dissociation \((\alpha) = \frac{i − 1}{n − 1}\)
\(\text{or, }0.23 = \frac{i − 1}{2 − 1}\)
\(\text{or, }i = 1.23\)
Now \(\Delta T = i × K_f × molality\)
⇒ \(\Delta T =1.23 × 1.86 × \frac{3 × 1000}{60 × 498.5}\)
⇒ \(\Delta T = \frac{1.23 × 93}{498.5}\)
∴ \(\Delta T = 0.229\)