What is the freezing point of a solution containing 8.1 g HBr in 100 g water, assuming that the acid to be 90% ionized? (Kf (H2O) = 1.86 K kg mol-1)

Question

What is the freezing point of a solution containing 8.1 g HBr in 100 g water, assuming that the acid to be 90% ionized? (Kf (H2O) = 1.86 K kg mol-1) (A) 0.85°C (B) -3.8°C (C) -0°C (D) -3.5°C

Tardigrade Admin · Accepted Answer

Consider the dissociation of 1 mol HBr. The degree of dissociation is 90%, i.e.,

α = 0.90

I ni t ia l co n c . F ina l co n c . H B r_{(a q)} 1 m o l 1 - 0.90 = 0.10 m o l \to H^{+} (a q) 0 0.90 m o l + B r^{-} (a q) 0 0.90 m o l

∴

Total number of particles

= 0.10 + 0.90 + 0.90

= 1.90 m o l

van’t Hoff factor

(i) = \frac{1.90}{1.00} = 1.90

Δ T_{f} = i K_{f} \frac{W _{B}}{M _{B} \cdot W _{A}} = \frac{1.90 \times 1.86 K k g m o l ^{- 1} \times 8.1 g}{81 g m o l ^{- 1} \times 100 \times 1 0 ^{- 3} k g}

= 3.5 K = 3. 5^{\circ} C

F.P. of solution = F.P. of solvent - depression in F.P.

= 0^{\circ} C - 3. 5^{\circ} C = - 3. 5^{\circ} C

Q. What is the freezing point of a solution containing 8.1 g HBr in 100 g water, assuming that the acid to be 90% ionized?
$(K_{f} (H_{2} O) = 1.86 K k g m o l^{- 1})$

$0.8 5^{\circ} C$

$- 3. 8^{\circ} C$

$- 0^{\circ} C$

$- 3. 5^{\circ} C$

Q. What is the freezing point of a solution containing 8.1 g HBr in 100 g water, assuming that the acid to be 90% ionized? (Kf​(H2​O)=1.86Kkgmol−1)

0.85∘C

−3.8∘C

−0∘C

−3.5∘C

Q. What is the freezing point of a solution containing 8.1 g HBr in 100 g water, assuming that the acid to be 90% ionized?
$(K_{f} (H_{2} O) = 1.86 K k g m o l^{- 1})$

$0.8 5^{\circ} C$

$- 3. 8^{\circ} C$

$- 0^{\circ} C$

$- 3. 5^{\circ} C$