Question

$0.6g$ of carbon was burnt in the air to form $C{O}_2$ The number of molecules of $C{O}_2$ introduced into the air will be:

• A. $8.01\times {10}^{23}$
• B. $3.01\times {10}^{22}$
• C. $6.02\times {10}^{23}$
• D. $3.01\times {10}^{23}$

Answer 1

Answer: (B)

$3.01\times {10}^{22}$

$C+O_2⟶C{O}_2$

$0.6g$ of carbon cations $=\frac{0.6}{12}$ moles of carbon.

as $1$ mole of $carbon$ gives $1$ mole of $C{O}_2$,

thus the no. of moles of $C{O}_2$ produced by $0.6$ of $C$ is $\frac{0.6}{12}$ moles.

$1$ mole $=6.022\times {10}^{23}$ molecule.

So, no. of molecules of $C{O}_2$ in $\frac{0.6}{12}$ moles of $C{O}_2$

$=\frac{0.6}{12}\times 6.022\times {10}^{23}$

$=3.011\times {10}^{23}$ molecule of $C{O}_2$ introduce in the air.

Hence, the correct option is B.