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From 200 mg of CO_(2), 10^(21)molecules are removed. How many moles of CO_(2) are left? |
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Answer» Solution :Total no. Of moles of `CO_(2) = ("WT.in g")/("MOL. Wt")` `=0.2/44 =0.00454` No. Of moles REMOVED `=(10^(21))/(6.022 XX 10^(23)) = 0.00166` No. Of moles of `CO_(2)` left `=0.00454 - 0.00166=0.00288` |
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