The density of a solution containing `13%` by mass of sulphuric acid is `1.09g//mL`. Calculate the molarity and normality of the solution

Explanation

Volume of 100g of the solution
`= (100)/(d)=(100)/(1.09)mL`
`=(100)/(1.09xx1000)` litre
`=(1)/(1.09xx10)` litre
Number of moles of `H_(2)SO_(4)` in 100g of the solution `= (13)/(98)`
Molarity `= ("No. of moles of "H_(2)SO_(4))/("Volume of soln. in litre")=(13)/(98)xx(1.09xx10)/(1)`
`=1.445M`
[Note : In solving such numbericals, the following formula can be applied :
Molarity `= (% "strength of soln." xx"density of soln. "xx10)/("Mol.mass")`
Similarly,
Normality `= (%"strength of soln." xx"densityof soln."xx10)/("Eq. mass")"]"`
We know that,
Normality = Molarity `xx n`
`= 1.445xx2[n=("Mol. mass")/("Eq. mass")=(98)/(49)=2]`
`=2.89N`.