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A small amount of CaCO(3) completely neu...

A small amount of `CaCO_(3)` completely neutralises 525 ml of `(N)/(10)` HCl and no acid is left at the end after converting all calcium chloride to `CaSO_(4)`. How much plaster paris `(CaSO_(4)(1)/(2)H_(2)O)` can be obtained

A

`1.916g`

B

`5.827g`

C

`7.53g`

D

`3.81g`

Text Solution

Verified by Experts

The correct Answer is:
D
No. of Meq of HCl = `525xx(1)/(10)=52.5`
No. of Meq of HCl = No. of mecq of `CaCl_(2)` =
No. of Meq of plaster of paris
So No. of Meq of plaster of paris = 52.5
`52.5=("wt of plaster of paris")/("GEW of plaste of paris")xx1000`
`52.5=(x)/(72.5)xx1000impliesx=3.81g`
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