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From 200 mg of `CO_(2),10^(21)` molecules are removed. How many moles of `CO_(2)` are left ? |
Answer» Step I : No. of molecules in 200 mg of `CO_(2)` Molar mass of `CO_(2)=12+32 = 44` g 44 g of `CO_(2)` contain molecules `= 6.022 xx 10^(23)` 200 mg or 0.2 g of `CO_(2)` contain molecules `= (6.022xx10^(23))/((44g))xx(0.2g) = 2.74 xx 10^(21)` Step II. No. of molecules of `CO_(2)` left Total no. of `CO_(2)` molecules `= 2.74 x 10^(21)` No. of `CO_(2)` molecules removed `= 10^(21)` No. of `CO_(2)` molecules left `= 2.74 xx 10^(21) - 10^(21) = 1.74 xx10^(21)` Step III. No. of moles of `CO_(2)` left `6.022 xx 10^(23)` molecules of `CO_(2)=1` mol `1.74xx10^(21)` molecules of `CO_(2)=(1 mol)/(6.022xx10^(23))xx1.74xx10^(21)=2.89xx10^(-3)` mol. |
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