If the density of a 1-ltre solution of `98% H_(2)SO_(4) (wt.//vol.)` is `1.88 g mol^(-1)`, the molality of the solution will be

Options

A. `13.13`
B. `10.10`
C. `11.11
D. `12.12`

Answer

3

Explanation

`98% H_(2)SO_(4) (wt.//vol.)` implies that every `100 mL` of solution contains `98 g` of `H_(2)SO_(4)`. Thus
`n_(H_(2)SO_(4))=(mass_(H_(2)SO_(4)))/(molar mass_(H_(2)SO_(4)))`
`=(98 g)/(98 g//mol)=1 mol`
Mass of solution =`(Volume)_("soln")xx("density")_("soln")`
`=(100 mL)(1.88 g mL^(-1))`
`=188 g`
Mass of solvent =(Mass of solution) - (Mass of solute)
`=(188 g) - (98 g)`
`=90 g`
Molality (m)`= n_(H_(2)SO_(4))/g_("solvent")xx(1000 g)/(kg)`
`=(1 mol)/(90 g)xx(1000 g)/(kg)`
`=11.11 m`