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.]

K_f(H₂O) = 1.85 K kg mol^–1

पर्याय

• 5

• 8

• 10

• 15

Answer 1

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 5.

Explanation:

Given, volume of acetic acid (v) = 1.2 mL

Density of acetic acid (d) = 1.02 g mL^–1

So, mass of acetic acid (m) = Density × volume

= 1.02 × 1.2

= 1.224 g

Now, molar mass of acetic acid = 60 g mol^–1

So, No. of moles of acetic acid = `"Mass"/"Molar mass"`

= `1.224/60`

= 0.0204

So, Moles = 0.0204 mole in 2L

So, Molality = 0.0102 `"mol"/"kg"`

∆T_f = i × K_f × m

0.0198 = i × 1.85 × 0.0102

i = `0.0198/(1.85 xx 0.0102)`

= 1.049

i ≈ 1.05

Since, α = `("i" - 1)/("n" - 1) = (1.05 - 1)/1` = 0.05

% α = 0.05 × 100%

= 5%

The percentage of dissociation of the acid is 5.