A bottle of commercial sulphuric acid (density = 1.787 gm/mL) is labelled 86 percent by weight. What is the molarity of the solution ? What volime of the acid is required to make it 1 litre of 0.2 M `H_(2)SO_(4)` ?

Answer

(i) 15.68 M (ii) 12.75 mL

Explanation

Step I. Molarity of solution
Mass of `H_(2)SO_(4)=86 g` , Mass of the solution = 100 g
Volume of 100 g of the solution `= ("Mass")/("Density") = (100)/(1.787) = 55.96 mL = 0.05596 L`
Molarity of solution (M) `= (("Mass of "H_(2)SO_(4))/("Molecular mass of "H_(2)SO_(4)))/("Volume of solution in mL"/(1000))=((86g))/(("98 g mol"^(-1))xx("0.05596 L"))`
`=15.68 "mol L"^(-1)=15.68 M`
Step II. Volume of solution required
`overset(("conc."))(M_(1)V_(1))-=overset(("dilute"))(N_(2)V_(2))`
`15.68xxV_(1)=0.2xx1000,V_(1)=(0.2xx1000)/(15.68)=12.75mL`.