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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
Molar mass `=(K_(f)xx1000xx w)/(Delta T_(f)xx W)=(1.86xx1000xx1.25)/(20xx1.1)`
`=105.68=105.7`.
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