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

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 ?

Text Solution

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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
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Knowledge Check

  • How much of NaOH is required to neutralise 1500 cm^(3) of 0.1 N HCl ? (Na = 23)

    A
    60 g
    B
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    C
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  • How much of NaOH is required to neutralise 1500cm^(3) of 0.1 N HCl ? (Na = 23)

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    B
    4 g
    C
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    D
    60 g

  • How much of NaOH is reuired to neutralise 1500 cm^(3) of 0.1 N HCl (Na=23)?

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    B
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    C
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    (d)60 g

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