Question

what volume of 2N and H2SO4 should be mixed so that the resultant solutions of 1L has normality = 3.

Answer 1

solution :-

N1V1 = N2V2 = N3V3 ,
WHERE ( V3 = V2+V1 )

LET US ASSUME THE VOLUME OF 2NH2SO4 IS X LITRE.

then the volume of 6NH2SO4 is ( 1 - x ) L

2 × x + 6 ( 1 - x ) = 3 ×1 ;

; 2x + 6 - 6x = 3

; x = 3/4 or 0.75 in litre