Commercially available sample of sulphuric acid is 15% `H_2SO_4` by weight (density=1.10g `mL^-1`) . Calculate the molarity of the solution.

Explanation

Let Mass of `15% H(2) SO_(4)` solution = 100 g
`:.` Mass of `H_(2) SO_(4) = 15 g`
Mass of Water `= 100 - 15 = 85 g = 0.085` kg
No . of moles of `H_(2) SO_(4) = (15 g)/( 98 g mol^(-1))`
= 0.153 mol
(Molar mass of `H_(2) SO_(4) = 98 g mol^(-1)`)
Molality ` = ("Moles of solute" (H_(2) SO_(4)))/("Mass of solvent (water) in kg")`
`= mol kg ^(-1)`
`(0.153 mol)/( 0.085 kg) = (153)/(85)`
=1.80 m