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