Find the molaity and molatity of a 15% solution w/w of `H_(2)SO_(4)("density of "H_(2)SO_(4)=1.02gcm^(-3))`.

Explanation

Calculation of molality of the slution,
Weight of `H_(2)SO_(4)` in 100 g of solution=15.0 g
Molar mass of `H_(2)SO_(4)=98 g mol^(-1)`
Molality of solution (m) `=("No.of gram moles of "H_(2)SO_(4))/("Mass of warter in kg")=(15g//98gmol^(-1))/(0.085 kg)`
`= 1.8(mol kg^(-1))=1.8m" Calculation of molarity of the solution. Molarity of solution (M)" = ("No. of gram moles moles of "H_(2)SO_(4))/("Volume of solution in litres")`
`"Weight of solution "=("Weight of solution")/("Density")=((100g))/((1.02g cm^(-3))=98.0cm^(3)`
No. of gram moles of `H_(2)SO_(4)=((15.0g))/((98gmol^(-1)))=0.153 mol`
`"Molarity of solution (M)"= ((0.153mol))/(((98.0)/1000dm^(3)))=1.56mol (dm^(3))^(-1)=1.56 M`