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When 1.80g glucose dissolved in 90g of H...

When `1.80g` glucose dissolved in `90g of H_(2)O`,the mole fraction of glucose is

A

`0.00399`

B

`0.00199`

C

`0.0199`

D

`0.998`

Text Solution

Verified by Experts

The correct Answer is:
B
Mole fraction of glucose `(n)/(n+N)=(0.01)/(0.01+5)`
`0.00199`
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