0.5 molal aqueous solution of a weak acid (HX) is 20 % ionised. If Kf for water is 1.86 K Kg mol-1, the lowering in freezing point of the solution is

Question

0.5 molal aqueous solution of a weak acid (HX) is 20 % ionised. If Kf for water is 1.86 K Kg mol-1, the lowering in freezing point of the solution is (A) - 1.12 K (B) 0.56 K (C) 1.12 K (D) - 0.56 K

Tardigrade Admin · Accepted Answer

20 % ionisation of weak acid <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/5d44d35c0926f1ad99559fd9df0a4910-.png" /> Total moles at equilibrium =1-α+α+α=1+α i =( text Total moles / text Initial moles )=(1+α/1)=1+0.2=1.2 We know that, lowcring in freczing point of the solutions (Δ T b ) = i K f m =1.2 × 1.86 K kg mol -1 × 0.5 =1.116 K ≈ 1.12 K

Q. 0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If Kf​ for water is 1.86KKgmol−1, the lowering in freezing point of the solution is

- 1.12 K

0.56 K

1.12 K

- 0.56 K

20% ionisation of weak acid Total moles at equilibrium =1−α+α+α=1+α i= Initial moles Total moles ​=11+α​=1+0.2=1.2 We know that, lowcring in freczing point of the solutions (ΔTb​)=iKf​m =1.2×1.86Kkgmol−1×0.5 =1.116K≈1.12K

Q. $0.5$ molal aqueous solution of a weak acid (HX) is $20%$ ionised. If $K_{f}$ for water is $1.86 K K g m o l^{- 1}$ , the lowering in freezing point of the solution is