Question

from 280 mg of CO, 10^21 molecules are removed then the number of remaining g molecule of CO will be

Answer 1

first calculate total no. of moles in 280mg of CO . it becomes 10×10^-3 moles. Now find total number of molecules in 10×10^-3 moles of CO by multiplying it with avagadro number. it gives 60.02×10^20 molecules. Now subtract the removed number of molecules(10^21) by 60.02×10^20 (6.002×10^21) molecules , it gives 5.002×10^21 molecules which is the remaining numer of molecules.
Now by unitary method,
280mg->60.02×10^20molecules
? ->5.002×10^21 molecules
cross multiply for the answer and the answer will be 233mg.
This implies that the remaining mass of CO is 233mg.

Answer 2

Answer:

8.34 x10 ^-3 moles.

Explanation:

28 g – 6.022x10 23 molecules.

280 x103 g contains = 6.022 x10 21 molecules,

6.022 x10 21 – 1x10 21 = 5.022 x10 21 molecules.

6.022 x10 23 molecules = 1 mole

5.022 x10 21 molecules = 8.34 x10 -3 moles.