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Standard VIII

Chemistry

Ways to Define Concentration

The freezing ...

Question

The freezing point (in $^{\circ} C$ ) of a solution containing 0.1 $g$ of $K_{3} [F e (C N)_{6}] (M o l . W t .329) i n 100 g w a t e r (K f = 1.86 K k g m o l^{- 1}$ is

- 1.2 \times 10^{- 2}

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- 5.7 \times 10^{- 2}

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- 5.7 \times 10^{- 3}

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- 2.3 \times 10^{- 2}

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Solution

The correct option is D $- 2.3 \times 10^{- 2}$
$T_{f}^{'} - T_{f}^{^{'}} = 0 - T_{f}^{^{'}} = 4 \times 1.86 \times \frac{\frac{0.1}{329}}{\frac{100}{1000}}$
$T_{f}^{^{'}} = - 4 \times 1.86 \times \frac{0.1}{329} \times \frac{1000}{100} = - 0.023$

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Similar questions

Q. The freezing point (in

^{o} C

) of a solution containing 0.1 g of

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

(molar mass 329) in 100 g of water

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

is :

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 |

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. Calculate the freezing point of a solution containing

60 g

of glucose (Molar mass

= 180 g m o l^{- 1}

) in

250 g

of water.

(

K f

of water

= 1.86 K k g m o l^{- 1}

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 (in $^{\circ} C$ ) of a solution containing 0.1 $g$ of $K_{3} [F e (C N)_{6}] (M o l . W t .329) i n 100 g w a t e r (K f = 1.86 K k g m o l^{- 1}$ is

The correct option is D $- 2.3 \times 10^{- 2}$
$T_{f}^{'} - T_{f}^{^{'}} = 0 - T_{f}^{^{'}} = 4 \times 1.86 \times \frac{\frac{0.1}{329}}{\frac{100}{1000}}$
$T_{f}^{^{'}} = - 4 \times 1.86 \times \frac{0.1}{329} \times \frac{1000}{100} = - 0.023$

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The freezing point (in ∘C) of a solution containing 0.1 g of K3[Fe(CN)6] (Mol.Wt.329) in 100 g water (Kf=1.86 K kg mol−1 is

The correct option is D −2.3×10−2T′f−T′f=0−T′f=4×1.86×0.13291001000 T′f=−4×1.86×0.1329×1000100=−0.023

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The freezing point (in $^{\circ} C$ ) of a solution containing 0.1 $g$ of $K_{3} [F e (C N)_{6}] (M o l . W t .329) i n 100 g w a t e r (K f = 1.86 K k g m o l^{- 1}$ is

The correct option is D $- 2.3 \times 10^{- 2}$
$T_{f}^{'} - T_{f}^{^{'}} = 0 - T_{f}^{^{'}} = 4 \times 1.86 \times \frac{\frac{0.1}{329}}{\frac{100}{1000}}$
$T_{f}^{^{'}} = - 4 \times 1.86 \times \frac{0.1}{329} \times \frac{1000}{100} = - 0.023$