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

`M = (W_(2) xx 1000)/(Mw_(2) xx V_(sol))`
`= (410.3 xx 1000)/(98 xx 1000) = 4.186`
`m = (W_(2) xx 1000)/(Mw_(2) xx (V xx d - W_(2)))`
`= (410.3 xx 1000)/(98 xx (1000 xx 1.243 - 410.3))`
`= (410.3 xx 1000)/(98 xx 832.7) = 5.027`