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A hydrocarbon has density `1.25 gL^(-1)`at `STP` . The hydrocarbon isA. `C_(2)H_(4)`B. `C_(2)H_(2)`C. `C_(2)H_(6)`D. `C_(3)H_(6)` |
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Answer» Correct Answer - A 1L of hydrocarbon at STP weighs `= 1.25 g` `22.4 L` of hydrocarbon at STP weigh `= 1.25 xx 22.4 = 5/4 xx 22.4 = 28.0 g mol^(-1)` Hydrocarbon will be `C_(2)H_(4) = 2xx 12 + 4 = 28 g mol^(-1)` (b) , (c ) and (d) are not possible. |
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