The density `("in" g mL^(-1))` of a `3.60 M` sulphuric acid solution that is `29% H_(2)SO_(4)` (Molar mass `=98 g mol^(-1)`) by mass will be:

Options

A. `1.22`
B. `1.45`
C. `1.64`
D. `1.88`

Answer

1

Explanation

`Density _("soln")=(mass_("soln"))/(volume_("soln"))`
`3.60 M` sulphuric acid solution means that 3.6 mole `H_(2)SO_(4)` is dissolved in every `1 L (1000 mL)` of solution. Thus
`mass_(H_(2)SO_(4))=(n_(H_(2)SO_(4)))("moles mass"_(H_(2)SO_(4)))`
`=(3.6 mol)(98 g mol^(-1))`
`=352.8 g`
By definition
`mass % H_(2)SO_(4)=(mass_(H_(2)SO_(4)))/(mass_(solution))xx100%`
Thus
`mass_(soln)=(mass H_(2)SO_(4))/(mass % H_(2)SO_(4))xx100%`
`=(352.8 g)/(29%)xx100%`
`=1216 g`
Substituting this result, we get
`"density"_(soln)=((1216 g)/(1000 mL))/("Density of soln")`
`=1.216 g mL^(-1)`
`=1.22 g mL^(-1)`