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 = 100g , Mass of `H_(2)SO_(4)` (solute) = 15 g
Mass of water (solvent ) = `100-15=85g=(85)/(1000)kg=0.085kg`
`"Mass of water (solvent)"=100-15=85g=(85)/(1000)kg=0.085kg`
`"Molar mass of "H_(2)SO_(4)="98 g mol"^(-1)" "therefore" "15gH_(2)SO_(4)=(15g)/(98"g mol"^(-1))=0.153"moles"`
`"Molality"=("No. of moles of solute")/("Mass of solvent in kg")=(0.153mol)/(0.085kg)=1.8mol 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)=98.04cm^(3)=(98.04)/(1000)L=0.09804L`
`"Molarity "=("No. of moles of the solute")/("Volume of solution in litres")=(0.153mol)/(0.09804L)=1.56mol L^(-1)=1.56M`