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From 200 mg of CO(2), 10^(21) molecules ...

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

Text Solution

Verified by Experts

`44 g of CO_(2) = 1 mol = 6.023 xx 10^(23)` molecules
`:. 6.023 xx 10^(23)` molecules `= 44 g of CO_(2)`
`10^(21)` molecules `= (40 xx 10^(21) xx 10^(3))/(6.023 xx 10^(23)) mg`
Weight of `CO_(2)` left `= 200 - 73.05 = 126.9 mg`
`(126.9)/(10^(3)) = 0.1269 g`
`44 g of CO_(2) = 1 mol`
`0.1269 g of CO_(2) = (1)/(44) xx 0.1269 = 0.0028 mol`
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