The freezing point of a solution prepared from 1.25 gm of a non-electrolyte and 20gm of water is 271.9 K. If molar depression constant is 1.86 K mole^-1, then molar mass of the solute will be

105.7

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106.7

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115.3

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93.9

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The correct option is A
105.7

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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:

The freezing point of a solution prepared from 1.25 gm of a non-electrolyte and 20gm of water is 271.9 K. If molar depression constant is 1.86 K mole^-1, then molar mass of the solute will be

1.00 g

of a non-electrolyte solute (molar mass

250 g m o l^{- 1}

) was dissolved in

51.2 g

of benzene. If the freezing point depression constant,

K_{f}

of benzene is

5.12 K k g m o l^{- 1}

, the freezing point of benzene will be lowered by:

Q. 1 g of non-electrolyte solute dissolved in 50g of benzene lowered the freezing point of benzene by 0.40K. The freezing point depression constant of benzene is 5.12 K kg/mol. Find the molar mass of the solute.

Q. Molal depression constant for water is

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

. The freezing point of a

0.05

molal solution of a non-electrolyte in water is:

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The freezing point of a solution prepared from 1.25 gm of a non-electrolyte and 20gm of water is 271.9 K. If molar depression constant is 1.86 K mole-1, then molar mass of the solute will be

The correct option is A 105.7

The freezing point of a solution prepared from 1.25 gm of a non-electrolyte and 20gm of water is 271.9 K. If molar depression constant is 1.86 K mole^-1, then molar mass of the solute will be

The correct option is A
105.7