What is the molarityk and molality of a 13% solution (by weight) of sulphric acid with a density of `1.02 mL^(-1)`? To what volume should `100 mL` of this acid be diluted in order to preapre a `1.5 N` solution?

Answer

A, B, C

Explanation

Molality `= (% "by weight" xx 10 xx d)/(Mw_(2))`
`= (13 xx 1.02 xx 10)/(98) = 1.35 M`
13% solution by wieght means `13 g` of solute is dissolved in `87 g` of solvent.
`m = (W_(2) xx 1000)/(Mw_(2) xx W_(1))`
`= (13)/(98) xx (1)/(87) xx 1000 = 1.52 m`
Normality = Molarity `xx 2` ('n' factor for `H_(2) SO_(4)`)
Normality `= 1.35 xx 2 = 2.70`
For dilution,
`N_(1)V_(1) = N_(2) V_(2)`
or `100 xx 2.70 = 1.5 xx V_(2)`
or `V_(2) = 180 mL`