The freezing point (in°C) of solution containing 0.1 g of K3[Fe(CN)6] (mol. wt. 329) in 100 g of water (Kf = 1.86 K kg mol-1) is

Question

The freezing point (in°C) of solution containing 0.1 g of K3[Fe(CN)6] (mol. wt. 329) in 100 g of water (Kf = 1.86 K kg mol-1) is (A) -2.3 × 10-2 (B) -5.7 × 10-2 (C) -5.7 × 10-3 (D) -1.2 × 10-2

Tardigrade Admin · Accepted Answer

van-t Hoff factor (i)= 4 3K+ [Fe(CN)6]3- Molality =(0.1/329) × (1000/100) =(1/329) ⇒ -Δ Tf =iKf m =4 × 1.86 × (1/329) =2.3 × 10-2 = Tf =-2.3 × 1-2 °C (As % ) freezing point of water is 0°C)

Q. The freezing point (in $^{\circ}$ C) of solution containing $0.1 g$ of $K_{3} [F e (CN)_{6}$ ] (mol. wt. $329)$ in $100 g$ of water $(K_{f} = 1.86 K k g m o l^{- 1})$ is

$- 2.3 \times 1 0^{- 2}$

$- 5.7 \times 1 0^{- 2}$

$- 5.7 \times 1 0^{- 3}$

$- 1.2 \times 1 0^{- 2}$

Q. The freezing point (in∘C) of solution containing 0.1g of K3​[Fe(CN)6​] (mol. wt. 329) in 100g of water (Kf​=1.86Kkgmol−1) is

−2.3×10−2

−5.7×10−2

−5.7×10−3

−1.2×10−2

van-t Hoff factor (i)=4{3K+[Fe(CN)6​]3− Molality =3290.1​×1001000​=3291​ ⇒−ΔTf​=iKf​m =4×1.86×3291​=2.3×10−2 =Tf​=−2.3×1−2∘C (As % ) freezing point of water is 0∘C)

Q. The freezing point (in $^{\circ}$ C) of solution containing $0.1 g$ of $K_{3} [F e (CN)_{6}$ ] (mol. wt. $329)$ in $100 g$ of water $(K_{f} = 1.86 K k g m o l^{- 1})$ is

$- 2.3 \times 1 0^{- 2}$

$- 5.7 \times 1 0^{- 2}$

$- 5.7 \times 1 0^{- 3}$

$- 1.2 \times 1 0^{- 2}$