1.

2N-HCl will have the sae molar concentration is

Answer»

`0.5 N-H_(2)SO_(4)`
`1.0N-H_(2)SO_(4)`
`2N-H_(2)SO_(4)`
`4N-H_(2)SO_(4)`

Solution :First 300 ML of 0.1 N HCL neutralises the entire amount of NaoH and 1/2 of `Na_(2)CO_(3)`.
As given, 25 ml of 0.2 N HCl neutralises remaining 1/2 of `Na_(2)CO_(3)`.
So number of milliequivalents of HCl=no. of
Milliequivalents of NaOH
`250xx0.1" N HCl"=250xx0.1" N NaOH"`
Which is equal to 1g of NaOH.


Discussion

No Comment Found

Related InterviewSolutions