22.7 mL of (N/10) `Na_(2)CO_(3)` solution neutralises 10.2 mL of a dilute `H_(2)SO_(4)` solution. The volume of water that must be added to 400 mL of this `H_(2)SO_(4)` solution in order to make it exactly N/10.A. 490.2 mLB. 890.2 mLC. 90.2 mLD. 290.2 mL

22.7 mL of (N/10) `Na_(2)CO_(3)` solution neutralises 10.2 mL of a dil

1.	22.7 mL of (N/10) `Na_(2)CO_(3)` solution neutralises 10.2 mL of a dilute `H_(2)SO_(4)` solution. The volume of water that must be added to 400 mL of this `H_(2)SO_(4)` solution in order to make it exactly N/10.A. 490.2 mLB. 890.2 mLC. 90.2 mLD. 290.2 mL
Answer» Correct Answer - A meq of `Na_(2)CO_(3)` = meq of `H_(2)SO_(4)` `(1)/(10)xx22.7=Nxx10.2` Normality = 0.2225 N `0.2225xx400=(1)/(10)xxV_(f)` or `V_(f)=890.2` mL `:. ` Volume of `H_(2)O` mixed = 890.2-400=490.2 mL

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22.7 mL of (N/10) `Na_(2)CO_(3)` solution neutralises 10.2 mL of a dilute `H_(2)SO_(4)` solution. The volume of water that must be added to 400 mL of this `H_(2)SO_(4)` solution in order to make it exactly N/10.A. 490.2 mLB. 890.2 mLC. 90.2 mLD. 290.2 mL

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