A small amount of `CaCO_(3)` completely neutralises 525 mL of 0.1 N HCl and no acid is left in the end. After converting all calcium chlorine to `CaSO_(4)`, how much plaster of Paris can be obtained ?

Explanation

525 mL of 0.1 N HCl=525 mL of 0.1 N `CaCl_(2)`
=525 mL of 0.1 N plaster of Paris
Molecular mass of plaster of Paris`=CaSO_(4). (1)/(2)H_(2)O=145`
Equivalent mass of plaster of Paris `=(145)/(2)=72.5`
Mass of plaster of Pairs in 525 mL of 0.1 N solution
`=(Nxx ExxV)/(1000)=(0.1xx72.5xx525)/(1000)`
=3.806 g