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`100 mL` of phosphine `(PH_(3))` on hearing forms phosphorous `(P)` and hydrogen `(H_(2))`. The volume change in the reaction isA. an increase of 50 mLB. an increase of 110 mLC. an increase of 150 mLD. a decrease of 50 mL |
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Answer» Correct Answer - A `underset(100mL)underset(2mL)(2PH_(3))(g)to2P(s)+underset(150mL)underset(3mL)(3H_(2))(g)` `:.` 100 mL of `PH_(3)` gives `H_(2)=150mL` Increase in volume `=150-100=50mL` |
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