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The molality of a sulphuric acid aqueous...

The molality of a sulphuric acid aqueous solution in which the mole fraction of water is 0.85 is

A

9.8

B

10.58

C

6.5

D

11.25

Text Solution

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The correct Answer is:
A
`(X_(H_(2)SO_(4)))/(X_(H_(2)O)) = (0.15)/(0.85) = (n_(H_(2)SO_(4)))/(n_(H_(2)O))`
`(n_(H_(2)SO_(4)))/(n_(H_(2)O)) = (0.15)/(0.85 xx 18)`
Molality `= (0.15 xx 1000)/(0.85 xx 18) = 9.8 M`
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