Question

If ${{10}^{21}}$ molecules are removed from 200 mg of $C{{O}_{2}}$ , the number of moles of $C{{O}_{2}}$ left is:
(A) $2.88\times {{10}^{-3}}$
(B) $28.8\times {{10}^{-3}}$
(C) $0.288\times {{10}^{-3}}$
(D) $1.66\times {{10}^{-3}}$

Answer 1

Hint: For solving this type of question, we must know the molecular mass of the given compound, the definition and the value of one mole.
Only one option of all the given options will be correct. As, the first three are interrelated, try to focus on them primarily.

Complete step by step solution:
Let us see the basic concepts required to solve the given illustration;
Molecular mass-
The molecular mass is the total mass of the compound consisting of different atoms. It is the summation of the product of the atomic masses of atoms to their subscripts in the molecule.
One mole-
One mole of substance has Avogadro’s number of molecules i.e. $6.022\times {{10}^{23}}$ .
Now, let us move forward;
Illustration-
Molecular mass of $C{{O}_{2}}$ $= 44 g$
Given mass of $C{{O}_{2}}$ $= 200 mg$ $= 0.2 g$
Thus, the number of moles can be given as;
No. of moles = $\dfrac{0.2}{44}=\dfrac{1}{220}$
We know, one mole of substance has Avogadro’s number of molecules i.e. $6.022\times {{10}^{23}}$ .
Thus,
Number of moles = $\dfrac{{{n}_{mol}}}{Av.no.}$
where,
${{n}_{mol}}$ = number of molecules
Av, No. = Avogadro’s number
So, number of molecules is given as;
Number of molecules = Avogadro’s number $\times $ number of moles
Number of molecules = $6.022\times {{10}^{23}}$$\times $$\dfrac{1}{220}=2.73\times {{10}^{21}}molecules$
Now, as ${{10}^{21}}$ molecules are removed;
Number of molecules left = $2.73\times {{10}^{21}}-{{10}^{21}}=1.73\times {{10}^{21}}molecules$
Using the same equation as above;
We get,
Number of moles = $\dfrac{{{n}_{mol}}}{Av.no.}$ =$\dfrac{1.73\times {{10}^{21}}}{6.022\times {{10}^{23}}}=2.88\times {{10}^{-3}}moles$

Therefore, option (A) is correct.

Note: Do note to use units properly. Here, there is a concept of Avogadro’s number which isn’t in fact new but, we need to have a proper and basic knowledge before using it in the illustrations as above.