The freezing point ( in ° C ) of a 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 a 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

(Δ TF) text solution =(Δ TF) text solvent -(Δ TF) text solvent with solute (Δ TF) text Solute =(. molality .× KF × i) =(0.1/329 × 100) × 1000 × 4 × 1.86 =0.022° C (Δ TF) text solvent =0° C So (Δ TF) text Solution =0-0.02232 ≅-2.3 × 10-2

Q. The freezing point ( in $^{\circ} C$ ) of a 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 a 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

(ΔTF​)solution ​=(ΔTF​)solvent ​−(ΔTF​)solvent with solute ​ (ΔTF​)Solute ​=( molality ×KF​×i) =329×1000.1​×1000×4×1.86 =0.022∘C (ΔTF​)solvent ​=0∘C So (ΔTF​)Solution ​=0−0.02232 ≅−2.3×10−2

Q. The freezing point ( in $^{\circ} C$ ) of a 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 :

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

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

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

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