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20 mL of a solution of sulphuric acid neutrlise 21.2 mL of 30 % solution of sodium carbonate. How much water should be added to 100 mL of this solution to bring down its strength to decinormal ? |
Answer» Step1. Calculation of normality of sulphuric acid solution Strength of `Na_(3)CO_(3)` solution `= 30 g//100 mL = 300 gL^(-1)` Normality of `Na_(2)CO_(3)` solution `= ("Strength")/("Equivalent mass")=((300 gL^(-1)))/((53"g equiv"^(-1)))` `=5.66 "equiv. L"^(-1) =5.66N` `N_(1)xxV_(1)-=N_(2)xxV_(2)or N_(1)=(N_(2)xxV_(2))/(V_(1))` `H_(2)SO_(4))" "(Na_(2)CO_(3))` `N_(1)=((5.66N)xx(21.2"mL"))/((20.0mL))=6.0N` Step II. Calculation of volume of water to be added Normality of `H_(2)SO_(4)` solution `= 6.0 N` Volume of `H_(2)SO_(4)` solution `= 100 mL` Normality of diluted `H_(2)SO_(4)` solution `= 0.1 N` Volume of diluted `H_(2)SO_(4)` solution may be calculated as : `underset(("Diluted"H_(2)SO_(4)))(N_(1)xxV_(1))=underset(("Concentrated" H_(2)SO_(4)))(N_(2)xxV_(2))` `V_(1)=(N_(2)xxV_(2))/(N_(1))=((6.0N)xx(100mL))/((0.1N))=6000mL` Volume of water to be added `= (6000-100)=5900` mL. |
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