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Standard XII

Chemistry

Raoult's Law

A solution of...

Question

A solution of 1.25g of non - electrolyte in 20g of water freezes at 271.94 K. If K_f = 1.86 K kg mol^-1,then the molecular weight of the solute will be:

179.79g mol^-1

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207.8 g mol^-1

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209.6 g mol^-1

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109.6 g/ mol^-1

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Solution

The correct option is D
109.6 g/ mol^-1

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Similar questions

Q. A solution of 1.25 g of a non - electrolyte in 20 g of water freezes at 271.94 K. If

K_{f}

= 1.86 K molality -1 then the molecular wt. of the solute is

A solution of 1.25g of non - electrolyte in 20g of water freezes at 271.94 K. If K_f = 1.86 K kg mol^-1, then the molecular weight of the solute will be:

Q. A solution of

1.25

g of a certain non-volatile substance in

20

g of water freezes at

271.94

K. Calculate the molecular mass of the solute

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

Q. The freezing point of a solution prepared from 1.25 g of non-electrolyte and 20 g of water is 271.9K. If the molar depression constant is

1.86 K {m o l}^{- 1}

, then molar mass of the solute will be:

40

g of a substance were dissolved in

450

g of water to give a solution which freezes at

- {0.8}^{\circ} C

. Calculate molecular weight of substance.

K_{f}

for water =

1.86

K kg mol-1.

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A solution of 1.25g of non - electrolyte in 20g of water freezes at 271.94 K. If Kf = 1.86 K kg mol-1,then the molecular weight of the solute will be:

The correct option is D 109.6 g/ mol-1

A solution of 1.25g of non - electrolyte in 20g of water freezes at 271.94 K. If K_f = 1.86 K kg mol^-1,then the molecular weight of the solute will be:

The correct option is D
109.6 g/ mol^-1