Question

what would be the normality of 0.2m of hno3​

Answer 1

using law of dilution ,

N1V1=N2V2

Normality = n x Molarity

total volume of solution = 400ml

normality of HNO3 = 0.2m

normality of H2SO4 = 2 x 0.2 = 0.4m

normality of NaOH= 0.1N

applying law of dilution ,

N x 400 = 0.1 x 100 + 0.4 x 200 - 0.1 x 100

N x 400 =80

N = 0.2

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Answer 2

Answer:

=0.6N

Given,

H3NO3 is a disaster acid as it contains two - OH ions

Thus, basically × molarity

= 3×0.2

= 0.6N

therefore, = 0.6N is a correct answer