100 mL of phosphine (PH(3)) on hearing f...

`100 mL` of phosphine `(PH_(3))` on hearing forms phosphorous `(P)` and hydrogen `(H_(2))`. The volume change in the reaction is

an increase of 50mL

an increase of 100mL

an increase of 150mL

an decrease of 50mL

Text Solution

Verified by Experts

The correct Answer is:

`underset(4mL)underset(4"mol")(4PH_(3)(g)) to P_(4)(s)+underset(6mL)underset(6"mol")(6H_(2)(g))`
`"Volume of "H_(2) "produced by 100mL "PH_(3)=(6)/(4)xx100=150mL`
Thus, there is an increase in an increase of 50mL

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