(a) What is the normality of a 96 per cent solution of `H_(2)SO_(4)` of specific gravity 1.84 ? (b) How many mL of 96 per cent sulphuric acid solution is necessary to prepare one litre 0.1 N `H_(2)SO_(4)` ? ( c) To what volume should 10 mL of 96 per cent `H_(2)SO_(4)` be diluted to prepure 2 N solution ?

(a) What is the normality of a 96 per cent solution of `H_(2)SO_(4)` o

1.	(a) What is the normality of a 96 per cent solution of `H_(2)SO_(4)` of specific gravity 1.84 ? (b) How many mL of 96 per cent sulphuric acid solution is necessary to prepare one litre 0.1 N `H_(2)SO_(4)` ? ( c) To what volume should 10 mL of 96 per cent `H_(2)SO_(4)` be diluted to prepure 2 N solution ?
Answer» Mass of 1 litre of `H_(2)SO_(4)` solution `="Vol."xx"Density"` `=1000xx1.84=1840 g` Mass of `H_(2)SO_(4)` present in one litre 96% `H_(2)SO_(4)` solution `= (96)/(100)xx1840=1766.4 g` Strength of `H_(2)SO_(4)` solution `=1766.4 g//L` (a) Normality `=("Strength")/("Eq. mass")=(1766.4)/(49)=36.05 N` (b) Let the volume taken be `V_(1)` mL Applying `" " N_(1)V_(1)=N_(2)V_(2)` `N_(1)=36.05 N, V_(1) = ?, N_(2)=(N)/(10), V_(2)=1000` mL `36.05xxV_(1)xx1000` So, `V_(1)=(1000)/(36.05xx10)=2.77` mL i.e., 2.77 mL of `H_(2)SO_(4)` is diluted to one litre. (c ) ` underset("Before dilution")(N_(B)V_(B))= underset("After dilution ")(N_(A)V_(A))` `10xx36.05=V_(A)xx2` `V_(A)=180.25` mL i.e., 10 mL of given `H_(2)SO_(4)` is diluted to 180.25 mL.

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