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 (N

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; ]

Canvas Classes · Accepted Answer

Strategy:
 Find the molality from the given mass and volume of acetic acid. Then calculate  from the freezing point depression and find the degree of dissociation .

Step 1: Calculate moles of acetic acid
- Volume = 1.2 mL, Density = 1.02 g/mL.
- Mass . Molar mass = 60 g/mol.

Step 2: Calculate molality ()
Solvent = 2.0 L  2.0 kg (since density ).

Step 3: Determine 
, .

Step 4: Solve for dissociaton
.

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