1.

From 200 mg of CO_(2),10^(21) molecules are removed. How many moles of CO_(2) are left ?

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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