1.

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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