The freezing point (in ° C ) of a solution containing 0.1 g of K 3[ Fe ( CN )6. (Mol. Wt. 329) in 100 g of water ( K f =1.86 K kg mol -1) is

Question

The freezing point (in ° C ) of a solution containing 0.1 g of K 3[ Fe ( CN )6. (Mol. Wt. 329) in 100 g of water ( K f =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

K 3[ Fe ( CN )6] arrow 3 K ++[ Fe ( CN )6]3- i =4 Δ T f = K f × i × ( m / M ) × (1000/ W ) =1.86 × 4 × (0.1/329) × (1000/100) =2.3 × 10-2 T f =-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

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

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

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

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

$K_{3} [F e (CN)_{6}] \to 3 K^{+} + [F e (CN)_{6}]^{3 -}$
$i = 4$
$Δ T_{f} = K_{f} \times i \times \frac{m}{M} \times \frac{1000}{W}$
$= 1.86 \times 4 \times \frac{0.1}{329} \times \frac{1000}{100}$
$= 2.3 \times 1 0^{- 2}$
$T_{f} = - 2.3 \times 1 0^{- 2}$

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