Correct Answer – D
`M = (“percentage” xx 10 xx d)/(GMW)`

Correct Answer – 1
`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)`

Correct Answer – D
3.6 M `H_(2)SO_(4)` means `3.6xx98=352.8g` of sulphuric acid are present in 1000 mL of the solution
29.0 g of acid are present in solution = 100 g
352.8 g of acid are present in solution
`=((100g)xx(352.8g))/((29g))=1216g`
Density of solution `= (“Mass of solution”)/(“Volume of solution”)`
`=(1216g)/(1000mL)=1.22gmL^(-1)`

Correct Answer – D
`M=(x xxd xx10)/(m_(B))`
`d=(Mxxm_(B))/(x xx10)=(3.6xx98)/(29xx10)~~1.22″g mL”^(-1)`]

Correct Answer – A