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According to the given Question,the reaction is

2HNO3+2H2O <---------------> 2HNO2+2H2O+O2

Given Normality of a dibasic acid =0.1N

Therefore Molarity = Normality/acidity

Acidity=2,,,,As it is a dibasic acid

Molarity of NO3 = 0.1/2 = 0.05M

Molarity of OH= 0.45/18 * 1000/100

by solving we get , Molarity of OH = 0.25M

Volume of NO3 = n/Molarity of NO3

n=2 (beacuse ,as per the reaction number of nitrate ions are 2)

Volume of NO3= 2/0.05

by solving we get ,V=40 Litre = 40000 ml

Therefore ,, Molarity of resulting mixture = M1V1+M2V2 /V1+V2

M= 0.05*40000+0.25*100 / 40000+100

=5*400+25 / 40100

=2000+25/40100

= 2025/40100

=0.05M

Molarity of the resultant solution=0.05M

Hope it helps!!!!!!!!