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The freezing point of a solution prepare...

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

A

105.7

B

106.7

C

115.3

D

93.9

Text Solution

Verified by Experts

The correct Answer is:
A
`DeltaT_(f) = (1000K_(f) xx W_(2))/(Mw_(2) xx W_(1))` , `[DeltaT=273-271.9 =1.1]`
`1.1=(1000 xx 1.86 xx 1.25)/(Mw_(2) xx 20)`
`Mw_(2) = 105.68`
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