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A solution of substance containing 1.05 ...

A solution of substance containing `1.05` g per 100 mL was found to be isotonic with `3%` glucose solution . The molecular mass of the substance is :

A

`31.5`

B

`6.3`

C

630

D

63

Text Solution

Verified by Experts

The correct Answer is:
D
For isotonic solution `pi_(1) = pi_(2) , C_(1) = C_(2) , n_(1) = n_(2)`
`(W_(1))/(M_(1)) =(W_(2))/(M_(2)) rArr (1.05)/M = 30/180 rArr M = (1.05 xx180)/3 = 63 `
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