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5.85 g of NaCL is dissolved in 90g of wa...

5.85 g of NaCL is dissolved in 90g of water. The mole fraction of the soloute is

A

0.01

B

0.1

C

0.2

D

0.0196

Text Solution

Verified by Experts

The correct Answer is:
D
Given, `w_(B) = 5.85 g rArr w_(A) = 90g`
`M_(B) = 58.5 m mol^(-1), M_(A) = 18 g mol^(-1)`
`therefore chi_(B) = (n_(B))/(n_(A) + n_(B)) = ((w_(B))/(M_(B)))/(w_(A)/(M_(A)) + (w_(B))/(M_(B))) = ((5.85)/(58.5))/((5.85)/(58.5)) + (90)/(18) = (0.1)/(5.1) = 0.0196`
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