1.

What volume of 36 N and 1N sulphuric acid must be mixed to get 1L of 6N sulphuric acid

Answer»

96.7 ML of 36 N + 903.3 ml of 1N
142.8 ml of 36 N + 857.2 ml of 1N
903.3 ml of 36 N + 96.7 ml of 1N
100 ml of 36 N + 900 ml of 1N

Solution :LET volume of 36 N `H_(2)SO_(4)=x` litre and volume of `1 N H_(2)SO_(4)=(1-x)` litre.
Both these acids on mixing should give 6 equivalents of `H_(2)SO_(4)`
`:.36x+(1-x)=6` or `35x=5`
or `x=(5)/(35)=1/7 L=(1000 ml)/(7)=142.8`ml
`:.` 142.8 ml of 36N `H_(2)SO_(4)` and857.2 ml of 1N `H_(2)SO_(4)` should be mixed.


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