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Enthalpy

The freezing ...

Question

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:

- {2.57}^{\circ} C

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- {0.28}^{\circ} C

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- {5.7}^{\circ} C

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- {1.12}^{\circ} C

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Solution

The correct option is D $- {1.12}^{\circ} C$
$K_{3} [F e (C N)_{6}] \to 3 K^{+} + {[F e (C N)_{6}]}_{6}^{- 3}$
$i = 1 + (n - 1) α; α = 100 %$
i = n = 4
$Δ T_{f} = i \times K_{f} \times m$
$m = m o l a l i t y = \frac{n_{s o l u t e}}{w_{s o l v e n t (K g)}} = \frac{\frac{10}{329}}{\frac{200}{1000}}$
$= \frac{10}{329} \times \frac{1000}{200} = \frac{50}{329}$
$= 0.15 = 1.5 \times 10^{- 1}$
$Δ T_{f} = 4 \times 1.86 \times 0.15 = 1.12$
$T_{f} = 0 - 1.12 = - {1.12}^{\circ} C$

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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

^{\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 |

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)

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The freezing point of a solution containing 10 g of K3[Fe(CN)6] (mol wt. 329) in 200 gram of water (Kf=1.86 K Kg mol−1) is:

The correct option is D −1.12∘ CK3[Fe(CN)6]→3K++[Fe(CN)6]−3 i=1+(n−1)α;α=100% i = n = 4 ΔTf=i×Kf×m m=molality=nsolutewsolvent(Kg)=103292001000 =10329×1000200=50329 =0.15=1.5×10−1 ΔTf=4×1.86×0.15=1.12 Tf=0−1.12=−1.12∘ C

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: