1.

Calculate the normality of HCl solution whose 500 ml is utillised to neutralise the 1500 ml of `(N)/(10)` NaOH solution.

Answer» According to law of equivalence
Equivalent of HCl - Equivalent of NaOH
`N_(1)V_(1)=N_(2)V_(2)`
`rArr N_(1)xx(500)/(1000)=(1)/(10)xx(1500)/(1000)`
`rArr N_(1)=(150)/(500)=(3)/(10)` .


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