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`4.4 g of CO_(2)` and `2.24 "litre" of H_(2)` at `STP` are mixed in a container. The total number of molecules present in the container will be:A. `6.022xx10^(23)`B. `1.2044xx10^(23)`C. `2 "mole"`D. `6.023xx10^(24)` |
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Answer» Correct Answer - B `44 gCO_(2)=N "molecules"` `:. 4.4 g CO_(2) = (N)/(10)` molecules `22.4` litre `H_(2)` at STP `=N` molecules `:. 2.24` litre `H_(2)` at STP `=(N)/(10)` molecules Thus total molecules `= (N)/(10) + (N)/(10) = (N)/(5)` |
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