Question

A solution contains `410.3 g` `H_(2) SO_(4)` per litre of the solution at `20^(@)C`. If the density `= 1.243 g mL^(-1)`, what will be its molality and molarity?

Answer 1

Mol. Mass of `H_(2)SO_(4)=98`
No. of moles of `H_(2)SO_(4)=(410.3)/(98)=4.186`
Molarity of `H_(2)SO_(4)` solution `= ("No. of moles of "H_(2)SO_(4))/("Volume of soln. in litre")`
`=(4.186)/(1)=4.186M`
Mass of 1 litre `H_(2)SO_(4)` solution `= 1000 xx 1.243=1243g`
Mass of water `=(1243-410.3)=832.7g=(832.7)/(1000)kg`
Molality of solution `= ("No. of moles of "H_(2)SO_(4))/("Mass of water in kg")`
`=(4.186)/(832.7)xx1000`
`=5.027m`.