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Calculate the normality of the solution obtained by mixing (iii) 100 cc of 0.1 M H_(2)SO_(4) with 100 cc of 0.1 M NaOH. |
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Answer» SOLUTION :100 cc of 0.1 M `H_(2)SO_(4)=100x0.1xx"2 meq , 100 cc of 0.1 M NAOH "=100xx0.1" meq = 10 meq"` 10 meq of NaOH will NEUTRALIZE 10 meq of `H_(2)SO_(4) therefore H_(2)SO_(4)` left after NEUTRALISATION = 10 meq Volume of the solution `=100+100="200 cc " therefore " Normality of "H_(2)SO_(4)" in the solution "=("10 meq")/("200 cc")="0.0 N"` |
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