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For the reaction, `N_(2)(g)+3H_(2)(g)to 2NH_(3)(g)`, if molecular masses of `NH_(3) " and " N_(2) " and " M_(1) " and" M_(2)`, their equivalent masses are `E_(1) " and" E_(2)`, then `(E_(1)-E_(2))` is :A. `(2M_(1)-M_(2))/(6)`B. `M_(1)-M_(2)`C. `3M_(1)-M_(2)`D. `M_(1)-3M_(2)` |
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Answer» Correct Answer - A Equivalent mass of `N_(2), i.e., E_(2)=(M_(2))/(6)` Equivalent mass of `NH_(3), i.e., E_(1)=(M_(1))/(3)` Then, `" " E_(1)-E_(2)=(M_(1))/(3)-(M_(2))/(6)=(2M_(1)-M_(2))/(6)` |
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