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

`15%` solution of `H_(2)SO_(4)` means 15 g of `H_(2)SO_(4)` are present in 100 g of the solution, i.e.,
Mass of `H_(2)SO_(4)` dissolved = 15 g , Mass of the solution = 100 g ,
Density of the solution `= "1.02 g/cm"^(3)" (Given)"`
Calculation of molality : Mass of solution = 100 g , Mass of `H_(2)SO_(4)` (solute) = 15 g
Mass of water (solvent) `=100-15=85 g =(85)/(1000)kg=0.085kg`
Molar mass of `H_(2)SO_(4)="98 g mol"^(-1)" "therefore" 15 g "H_(2)SO_(4)=(15g)/("98 g mol"^(-1))="0.153 moles"`
`"Molality"=("No. of moles of solute")/("Mass of solvent in kg")=("0.153 mol")/("0.085 kg")="1.8 mol kg"^(-1)=1.8m`
Calculation of molarity : `"15 g of "H_(2)SO_(4)="0.153 moles (calculated above)"`
`"Volume of solution "=("Mass of solution")/("Density of solution")=(100)/(1.02)="98.04 cm"^(3)=(98.04)/(1000)L=0.09804L`
`"Molarity"=("No. of moles of the solute")/("Volume of solution in litres")=("0.153 mol")/("0.09804 L")="1.56 mol L"^(-1)="1.56 M"`