35.4 mL of HCl is required for the neutralization of a solution containing 0.275 g of sodium hydroxide. The normality of hydrochloric acid is

Options

A. 0.97 N
B. 0.142 N
C. 0.194 N
D. 0.244 N

Answer

C

Explanation

We know that 1g equivalent weight of
NaOH = 40 g
`therefore` 40 g of NaOH = 1 g eq. Of NaOH
`therefore` 0.275 g of NaOH = `(1)/(40)xx0.275` eq.
`=(1)/(40)xx0.275xx1000=6.88` meq.
`therefore underset((HCl))(N_(1)V_(1))=underset((NaOH))(N_(2)V_(2))`
`N_(1)xx35.4=6.88` (`because` meq = NV)
`N_(1)=0.194`.