1.

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

Answer»

0.97 N
0.142 N
0.194 N
0.244 N

Solution :We KNOW that 1 G EQUIVALENT weight of
NAOH = 40 g
`therefore` 40 g of NaOH = 1 g eq. of NaOH
`therefore 0.275 g` of `NaOH =(1)/(40xx0.275 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`.


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