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The term mole first used by Ostwald in `1896` refers for the ratio of mass of a substance in `g` and its molecular weight. `1 "mole"` of a gaseous compound occupies `22.4 "litre"` at `NTP` and contains `6.023xx10^(23)` molecules of gas. The volume of air needed to burning `12 g` carbon completely at `STP` is:A. `22.4 "litre"`B. `112 "litre"`C. `44.8 "litre"`D. `50 "litre"` |
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Answer» Correct Answer - B `C+O_(2)rarrCO_(2)` `12 g` carbon needs `22.4 "litre air" 22.4 "litre" O_(2)` or `(22.4xx100)/(20)"litre air"=112 "litre air"` |
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