1.

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

Answer»

9.8
10.58
6.5
11.25

Solution :`(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`


Discussion

No Comment Found

Related InterviewSolutions