Question

24. The normality of solution obtained by mixing
100 ml of 0.2 M H2SO4 with 100 ml of 0.2 M
NaOH is
(1) 0.1
(2) 0.2
(3) 0.5
(4) 0.3
​

Answer 1

Answer:normality of the resultant solution=N1V1+ N2V2/(VI+V2)

N1=M1* N-FACTOR=0.2*2=0.4      V1=100

N2= M2*N-FACTOR=0.2*1=0.2       V2=100

RESULTANT NORMALITY=0.4*100+0.2*100/(100+100)

                                           =0.3

Explanation:IF IT HELPS MARK AS BRAINLIEST

Answer 2

Normality of 0.2M H2SO4 =0.2 X 2=0.4 Normality of 0.2M NaOH =0.2 x 1= 0.2 N

No.of equivalent of H2SO4 present in 100 ml. = 0.1. X0.4 =0.04

No.of equivalent of NaOH present in 100 ml. = 0.1 x 0.2 =0.02

No.of equivalent of H2SO4 present in excess. =0.02

100ml.of 0.4N H2SO4 contains =0.4 normal

100 ml.of 0.2N NaOH contains =0.2 normal

So the resulting solution contains =0.4–0.2= 0.2Normal

200 ml.of solution contains 0.2 equivalent of H2SO4

So the normality =0.2/2=0.1 N

The normality of the resulting solution will be 0.1N of H2SO4

Please mark as brainlist...