5mL of `N HCI, 20 mL` of `N//2 H_(2)SO_(4)` and 30 mL of `N//3 HNO_(3)` are mixed together and volume made to one litre. The normality of the resulting solution is

Options

A. `(N)/(5)`
B. `(N)/(10)`
C. `(N)/(20)`
D. `(N)/(40)`

Answer

D

Explanation

For `HCI`,
`N_(1)V_(1) = N_(2)V_(2)`
`N xx 5 = N_(2) xx 1000`
`N_(2) = 5 xx 10^(-3) N =0.5 xx 10^(-2)N`
For `H_(2)SO_(4)`,
`N_(1)V_(1) = N_(2)V_(2)`
`(N)/(2) xx 20 = N_(2) xx 1000`
`N_(2) = 10^(-2)N`
For `HNO_(3)`,
`N_(1)V_(1) = N_(2)V_(2)`
`(N)/(3) xx 30 = N_(2) xx 1000`
`N_(2) = 10^(-2) N`
Total normality
`= N(HCI) + N(H_(2)SO_(4)) + N (HNO_(3))`
`= (0.5 xx 10^(-2) +106(-2) +10^(-2)) N`
`= 2.5 xx 10^(-2)N`
`(25)/(1000) N = (N)/(40)`