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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» Correct Answer - (i) 15.68 M (ii) 12.75 mL 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`. |
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