Volume of `0.6` M NaOH required to neutralise `30cm^(3)" of "0.4M` HCI is

Options

A. `40 cm^(3)`
B. `30 cm^(3)`
C. `20 cm^(3)`
D. ` 10 cm^(3)`

Answer

C

Explanation

Normality = molarity × basicity or acidity (for HCl)
`N_(2) = 0.4 xx 1 = 0.4 N` basicity =1 (for NaOH acidity =1)
`N_(1) = 0.6 xx 1 = 0.6 N , V_(1) = ? V_(2) = 30 cm^(3)`
From the equation ,`N_(1)V_(1) = N_(2)V_(2)`
`0.6 xx V_(1) = 0.4 xx 30`
`V_(1) = (0.4 xx 30)/(0.6) = 20 cm^(3)`