JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
\(1.2\, mL\) 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 \(0.0198^{\circ} C\). 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 \(60 \,g\, mol ^{-1}\) \(\left.K _{ f }\left( H _{2} O \right)=1.85\, K\, kg\, mol ^{-1}\right]\)
- A \(50\)
- B \(5\)
- C \(45\)
- D \(24\)
Answer & Solution
Correct Answer
(B) \(5\)
Step-by-step Solution
Detailed explanation
\(M = d \times V =1.02 \times 1.2=1.224\, gm\) Moles of acetic acid \(=0.0204\) \(moles\) in \(2\, L\) So molality \(=0.0102 \,mol / kg\) Now \(\Delta T _{ f }= i \times K _{ f } \times M\) \(i=1+\alpha\) for acetic acid \(0.0198=(1+\alpha) \times 1.85 \times 0.0102\)…
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