1.

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

Answer»

`100cm^(3)`
`150CM^(3)`
`250cm^(3)`
`200CM^(3)`

SOLUTION :Given, N=0.1, w=1.0 gm, equivalent weight=53,V=?
`V=(wxx1000)/(Eq.wt.xxN)=(1xx1000)/(53xx0.1)=188.67cm^(3) approx 200cm^(3)`


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