Volume of `100g` of the solution
`=(100)/(d) = (100)/(1.09)mL = (100)/(1.09 xx 1000)` litre `= (1)/(1.09 xx 10)` litre
Number of moles of `H_(2)SO_(4)` in `100 g` of the solution `= (13)/(98)`
Molarity `= ("No. of moles of" H_(2)SO_(4))/("Volume of solution, in litre") = (13)/(98) xx (1.09 xx 10)/(1) = 1.445 M`
Note: In solving such numericals, the following formula can be applied:
Molarity `=(%"strength of solution" xx "density of solution" xx 10)/("Mol. mass")`
Similarly,
Normally `=(%"strength of solution" xx "density of solution" xx 10)/("Eq. mass")`
We know that,
Normality = Molarity `xx n =1.445 xx2 [n =(Mol. mass)/(Eq. mass) =(98)/(49) =2] = 2.89 N`
