Calculate the molarity of `9.8%` (w/w) solution of `H_(2) SO_(4)` , if the density of the solution is `1.02 g mL^(-1)` . [ Molar mass of `H_(2) SO_(4) = 98 g mol ^(-1)`]

Explanation

`9.8%` solutions of `H_(2)SO_(4)` means 9.8 g of `H_(2) SO_(4)` is present in 100 g of the solution .
`:.` Mass of solution = 100 g
Density of solution `= 102 g m L^(-1)`
Moles of `H_(2) SO_(4) = 9.8 // 98 g mol^(-1)` = 0.1 mol
Volume of solution `= (100g)/(102 g mL^(-1)) = (100)/(102) ` mL or 0.09804 L
Molarity `= ("Number of moles of solute ")/("Volume of solution (in L)")= (0.1)/(0.09804)`
= 102 M