100 mL of `PH_(3)` on decomposition produced phosphorus and hydrogen. The change in volume is
A. 50 mL increase
B. 500 ml decrease
C. 900 mL decrease
D. nil
A. 50 mL increase
B. 500 ml decrease
C. 900 mL decrease
D. nil
Correct Answer – a
`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
Correct Answer – A
`underset(2vol)(2PH_(3))tounderset(3vol)(3H_(2))+2P`
2 vol of `PH_(3)` give 3 vol of `H_(2)`
`:.` 100 mL of `PH_(3)` gives 150 mL of `H_(2)`
`:.` Increases in volume = 150 mL – 100 mL
= 50 mL