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The volume occupied by `7.40 g of CO_(2)` (at `STP`) isA. `3.8 L`B. `4.5 L`C. `5.6 L`D. `2.9 L` |
Answer» Correct Answer - A Since `1 mol` of an ideal gas occupies `22.4 L at STP`, we have `V_(CO_(2)) = (n_(CO_(2))) (22.4 (L)/(mol))` `= ((7.40 g)/(44 g mol^(-1))) (22.4 L mol^(-1))` `= 3.76 L` |
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