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Crystalline and Amorphous Solid

The freezing ...

Question

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

- 2.3 \times 10^{- 3}

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5.7 \times 10^{- 2}

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5.7 \times 10^{- 3}

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- 1.20 \times 10^{- 2}

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Solution

The correct option is A $- 2.3 \times 10^{- 3}$
$Δ T_{f} = i K_{f} m = 4 \times 1.86 \times \frac{0.1}{329} \times \frac{1000}{100}$
=- 0.0023
for $K_{3} [F e (C N)_{6}] ⇌ 3 K^{\oplus} + F e (C N)_{6}^{3 -}$
i =4

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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 (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 (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 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:

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:

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The freezing point (∘C) of a solution containing 0.1 g of K3[Fe(CN)6] (molecular weight 329) in 100 g of water (Kf=1.86Kkgmol−1)

The correct option is A −2.3×10−3ΔTf=iKfm=4×1.86×0.1329×1000100 =- 0.0023 for K3[Fe(CN)6]⇌3K⊕+Fe(CN)3−6 i =4

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

The correct option is A $- 2.3 \times 10^{- 3}$
$Δ T_{f} = i K_{f} m = 4 \times 1.86 \times \frac{0.1}{329} \times \frac{1000}{100}$
=- 0.0023
for $K_{3} [F e (C N)_{6}] ⇌ 3 K^{\oplus} + F e (C N)_{6}^{3 -}$
i =4