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Solution
Here
normality N1 of HCl =?
Volume V1 of HCl =500mL
Volume V2 of NaOH = 1500mL
Normality N2 of NaOH = 1/10N
using formula
N1V1 =N2V2
N1.500 = 1/10.1500
N1= 0.3 N
hence normality of HCl =0.3N
Here
normality N1 of HCl =?
Volume V1 of HCl =500mL
Volume V2 of NaOH = 1500mL
Normality N2 of NaOH = 1/10N
using formula
N1V1 =N2V2
N1.500 = 1/10.1500
N1= 0.3 N
hence normality of HCl =0.3N