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The molality of a solute in a solvent (m...

The molality of a solute in a solvent (molar mass, `20g mol^(-1))` is `0.1 mol kg^(-1)`.
What is the mole fraction of the solute?

A

`0.1/50`

B

`0.1/50.1`

C

`0.1/(20+0.1)`

D

`0.1/20`

Text Solution

Verified by Experts

The correct Answer is:
2
A `0.1 m` solution contains `0.1` mole of solution in every `1 kg (1000 g)` of solvent. Since solution is a homogenous mixture, we can consider any size of solution to calculate its concentration. Let's pick an arbitray amount of the solvent that will make the calculations easy, say `1000 g`. Thus
`n_(solute)=0.1`
`n_(solvent)=(mass_(solvent))/(molar mass_(solvent))`
`=(1000 g)/(20 g mol^(-1))=50 mol`
Now `chi_(solute)=n_(solute)/(n_(solute)+n_(solvent))`
`=0.1/(0.1+50)`
`=0.1/50.1`
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