Question

If $1.8\times {10}^{19}$ molecules of $N{O}_2$ are removed from $10mg$ of $N{O}_2$, how many moles of $N{O}_2$ are left?

• A. $3.87\times {10}^4$
• B. $1.87\times {10}^{-4}$
• C. $1.87\times {10}^4$
• D. $2.87\times {10}^{-4}$

Answer 1

Answer: (B)

$1.87\times {10}^{-4}$

First, let's find out the number of molecules present in 10 mg of $N{O}_2$

The molecular weight of $N{O}_2=14$ (atomic weight of $N$ )$+2\times 16$ (atomic weight of Oxygen)

The molecular weight of $N{O}_2=46g$

From Avogadro's rule, 1 mole of the $N{O}_2$ contains $6.022\times {10}^{23}molecules$

$\Rightarrow$ 46g of $N{O}_2$ contains $6.022\times {10}^{23}molecules$

$ 1 g of $N{O}_2$ contians $\frac{6.022\times {10}^{23}}{46g}=0.13\times {10}^{23}=0.13\times {10}^{23}molecules$

$10mg=10\times {10}^{-3}g=0.13\times {10}^{23}\times 10\times {10}^{-3}=0.13\times {10}^{21}=13\times {10}^{19}molecules$

$\therefore$ 10mg of $N{O}_2$ contains $13\times {10}^{19}$molecules

Given that $1.8\times {10}^{19}$ molecules were removed from 10 mg of $N{O}_2$

$\Rightarrow$ $13\times {10}^{19}-1.8\times {10}^{19}=(13-1.8)\times {10}^{19}=\mathrm{11.2.}\times {10}^{19}molecules$

Now we need to convert these molecules to number of moles

If $6.022\times {10}^{23}$ molecules is 1 mole (By Avogadro's rule)

$11.2\times {10}^{19}$ moelcules contians $\frac{\text{1 mole}}{6.022\times {10}^{23}\text{mloecules}}\times 11.2\times {10}^{19}\text{molecules}$

$=1.87\times {10}^{-4}moles$