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100 mL of `PH_(3)` on decomposition produced phosphorus and hydrogen. The change in volume is

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`4PH_(3)(g) overset("Heat")toP_(4)(s)+6H_(2)(g)`
Volume of `H_(2)` formed by decomposition of 100mL `PH_(3)`
`=(6)/(4)xx100=150mL`
Thus, change in volume=(150-100)=50mL

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