Calculate the molality of 1 litre solution of 93% `H_2SO_4` (weight/volume). The density of the solution is `1.84 g ml^(-1)`.

Answer

10.42

Explanation

Given: 93% `H_(2)SO_(4)` solution (w/v) , 93g `H_(2)SO_(4)` is present in `rarr 100` ml of solution
100 ml solution `rArr` 184 gm solution (d = 1.84 gm/cc)
`:.` 184 – 93 = 91 gm solvent has 93 gm `H_(2)SO_(4) = (93)/(98)` moles `H_(2)SO_(4)`
Hence, 1000 gm solvent has `= (93)/(98) xx (1000)/(91) = 10.42` molal