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What is the volume of 0.1 N HCl required...

What is the volume of 0.1 N HCl required to react completely with 1.0 g of pure calcium carbonate? (Ca = 40, C = 12 and 0 = 16)

A

`100 cm^3`

B

`150 cm^3`

C

`250 cm^3`

D

`200 cm^3`

Text Solution

Verified by Experts

The correct Answer is:
D
`W= 1.0 g E.W = 50 (Ca CO_3)`
` V= ? , N= 0.1 N (HCI)`
As ` ( W)/( E.W ) = (V xx N ) /( 1000 ) impliesV = ( 1 xx 1000 )/( 50 xx 0.1 ) = 200 cm^3`
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