What is the volume of 0.1 N HCl required to react completely with 1.0 g of pure calcium carbonate
Answers
Answered by
1
Answer:
200cm3
Explanation:
N=0.1, w=1.0gm, Equivalent weight CaCO
3
=50, V
HCl
=?
Using the law of equivalence:
n
eq
=N×V⇒V=
N
n
eq
V=
Eq.wt.×N
w×1000
=
50×0.1
1×1000
=200cm
3
Answered by
1
Answer:
N = 0.1
Normality, N = W ×1000/ M × V
Where, W = equivalent weight of CaCO3=100/2 = 50
M = Molar mass of CaCO3 = 100
V = Volume of HCl required =V
Therefore,
0.1 = 1 ×1000/ 50× V
V = 200ml
hope this will help you.
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