A solution containing 0.85 g of Z n C l2 in 125.0 g of water freezes at- 0.23° C. The apparent degree of dissociation of the salt is (Kf. for water =1.86 K kg mol -1, atomic mass: Zn =65.3 and Cl =35.5 )

Question

A solution containing 0.85 g of Z n C l2 in 125.0 g of water freezes at- 0.23° C. The apparent degree of dissociation of the salt is (Kf. for water =1.86 K kg mol -1, atomic mass: Zn =65.3 and Cl =35.5 ) (A) 1.36% (B) 73.5% (C) 7.35% (D) 2.47%

Tardigrade Admin · Accepted Answer

Mol. w t=(kf × w × 1000/Δ Tf × W) =(1.86 × 0.85 × 1000/0.23 × 125) ≈ 55 gm Where w=0.85 g W=125 g Δ Tf=0° C-(-23° C)=23° C Now, i=(M text normal /M text observed )=(136.3/55)=2.47 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/c19884396de662cdfc56b38f53900c92-.png" /> Van't Hoff factor (i) =(1-α+α+2 α/1)=2.47 ∴ α=0.735=73.5 %

Q. A solution containing 0.85g of ZnCl2​ in 125.0g of water freezes at- 0.23∘C. The apparent degree of dissociation of the salt is (Kf​ for water =1.86Kkgmol−1, atomic mass: Zn=65.3 and Cl=35.5 )

1.36%

73.5%

7.35%

2.47%

Mol. wt=ΔTf​×Wkf​×w×1000​ =0.23×1251.86×0.85×1000​≈55gm Where w=0.85g W=125g ΔTf​=0∘C−(−23∘C)=23∘C Now, i=Mobserved ​Mnormal ​​=55136.3​=2.47 Van't Hoff factor (i) =11−α+α+2α​=2.47 ∴α=0.735=73.5%

Q. A solution containing $0.85 g$ of $Z n C l_{2}$ in $125.0 g$ of water freezes at- $0.2 3^{\circ} C$ . The apparent degree of dissociation of the salt is $(K_{f}$ for water $= 1.86 K k g m o l^{- 1}$ , atomic mass: $Z n = 65.3$ and $Cl = 35.5$ )