A 0.2 molal aqueous solution of a weak acid (HX) is 20 % ionised. The freezing point (in degree celsius) of this solution is (Given k f =1.86° C kg mol -1 for water) :

Question

A 0.2 molal aqueous solution of a weak acid (HX) is 20 % ionised. The freezing point (in degree celsius) of this solution is (Given k f =1.86° C kg mol -1 for water) : (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Δ T f =0.2 × 1.2 × 1.86=0.45 therefore, boiling point =-0.45° C

Q. A $0.2$ molal aqueous solution of a weak acid $(H X)$ is $20%$ ionised. The freezing point (in degree celsius) of this solution is (Given $k_{f} = 1.8 6^{\circ} C k g m o l^{- 1}$ for water) :

$Δ T_{f} = 0.2 \times 1.2 \times 1.86 = 0.45$
therefore, boiling point $= - 0.4 5^{\circ} C$

Q. A 0.2 molal aqueous solution of a weak acid (HX) is 20% ionised. The freezing point (in degree celsius) of this solution is (Given kf​=1.86∘Ckgmol−1 for water) :

ΔTf​=0.2×1.2×1.86=0.45 therefore, boiling point =−0.45∘C

Q. A $0.2$ molal aqueous solution of a weak acid $(H X)$ is $20%$ ionised. The freezing point (in degree celsius) of this solution is (Given $k_{f} = 1.8 6^{\circ} C k g m o l^{- 1}$ for water) :

$Δ T_{f} = 0.2 \times 1.2 \times 1.86 = 0.45$
therefore, boiling point $= - 0.4 5^{\circ} C$