A solution contians 510 g of of sulphuric acid per litre at `25 ^(@) C` Calculated the normality and molarity of the solution.

Explanation

Normality = ` ( " Number of gram equivalents of solute") / ( " Volume of solution is litres ") `
Gram equivalent =` ("Mass ")/( "Eq mass ") `
` "Equivalent Mass of H_2SO_4 = ( " Molar mass ")/( "Basicity ") `
` " " = ( 98)/( 2) = 49 `
` "Equivalencs of "H_2SO_4 = ( 510 g) / ( 49 ) = 10.408`
Normality `= ( 10.408)/( 1) = 10.40N`
(ii) Molariry = `(" No , of moles of solute")/( " Volume of solution in litre ") `
` " No , of moles " = ("Mass ") /( "Molecular mass ") `
` = ( 510)/( 98)= 5.20 ` moles
Molarity ` = (" No , of moles of solute ")/(" Vol . of solution in 1 lit") `
` = ( 5.20)/( 1) = 5.20 M`