1.

What is the volume of 0.1 N HCl required to react completely with 1.0 g of pure calcium carbonate​

Answer»

:- N=0.1, w=1.0gm, EQUIVALENT weight CaCo3 = 50, VHCl =?✿ SOLUTION :- USING the law of EQUIVALENCE: Neq =N×V⇒V= Neq/NV= w×1000 / EQ.Wt.×N = 1×1000/ 50×0.1 = 200CM3



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