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

Weight of solute = 15 g `(H_2SO_4)`
Weight of Solvent = 100 g `(H_2O)`
Grani mole of Solute `(H_2SO_4)= 15/98"mol "` (Molar mass of `H_2SO_4=98g`)
Volume = `("Mass")/("Density")`
` = (100)/(1.10) =90.9` mL
Molarity = `("Moles of" H_2SO_4)/("Volume") xx1000`
`= 15/98xx(1000)/(90.9)= 1.68 M`