The freezing point (in $^{\circ}$ C) of solution containing 0.1g of $K_{3} [F e (C N)_{6}]$ (mol.wt. 329) in 100g of water ( $K_{f} = 1.86 K k g m o l^{- 1}$ ) is $x \times 10^{- 2}$ . Find the value of $| x |$ .

Open in App

Solution

Van't hoff factor (i) $= 4 {3 K^{+} + {[F e (C N)_{6}]}_{6}^{3 -}}$
Molality, $m = \frac{0.1}{329} \times \frac{1000}{100} = \frac{1}{329}$
$\Rightarrow Δ T_{f} = i K_{f} . m = 4 \times 1.86 \times \frac{1}{329} = 2.3 \times 10^{- 2}$
$\Rightarrow T - f = - 2.3 \times 10^{- 2}^{\circ} C$

Suggest Corrections

Similar questions

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

Q. The freezing point

(^{\circ} C)

of a solution containing 0.1 g of

K_{3} [F e (C N)_{6}]

(molecular weight 329) in 100 g of water

(K_{f} = 1.86 K k g m o l^{- 1})

Q. The freezing point (in

^{o} C)

of a solution containing 0.1g of

K_{3} [F e (C N)_{6}]

(Mol. Wt. 329) in 100

g

of water

(K_{f} = 1.86 K

kg mol

^{- 1})

is:

Q. The freezing point (in

^{\circ} C

) of a solution containing 0.1 g of

K_{3} [F e (C N)_{6}] (m o l w t = 329)

in 100 g of water

(k f = 1.86 K k g / m o l)

Q. The freezing point of a solution containing 10 g of

K_{3} [F e (C N)_{6}]

(mol wt. 329) in 200 gram of water

(K_{f} = 1.86 K K g m o l^{- 1})

is:

Join BYJU'S Learning Program

The freezing point (in ∘ C) of solution containing 0.1g of K3[Fe(CN)6] (mol.wt. 329) in 100g of water (Kf=1.86 Kkgmol−1) is x×10−2. Find the value of |x|.

Van't hoff factor (i)=4{3K++[Fe(CN)6]3−} Molality, m=0.1329×1000100=1329 ⇒ΔTf=iKf.m=4×1.86×1329=2.3×10−2 ⇒T−f=−2.3×10−2 ∘C

The freezing point (in $^{\circ}$ C) of solution containing 0.1g of $K_{3} [F e (C N)_{6}]$ (mol.wt. 329) in 100g of water ( $K_{f} = 1.86 K k g m o l^{- 1}$ ) is $x \times 10^{- 2}$ . Find the value of $| x |$ .

Van't hoff factor (i) $= 4 {3 K^{+} + {[F e (C N)_{6}]}_{6}^{3 -}}$
Molality, $m = \frac{0.1}{329} \times \frac{1000}{100} = \frac{1}{329}$
$\Rightarrow Δ T_{f} = i K_{f} . m = 4 \times 1.86 \times \frac{1}{329} = 2.3 \times 10^{- 2}$
$\Rightarrow T - f = - 2.3 \times 10^{- 2}^{\circ} C$