Question

From $200$ mg of $C{O}_2$, ${10}^{21}$ molecules are removed. How many moles(in order of ${10}^{-3}$) of $C{O}_2$ are left?

[Round off to nearest integer]

Answer 1

$(12+32)$ g of $C{O}_2\to 6.02\times {10}^{23}$ molecules

$\Rightarrow 200mg$ of $C{O}_2\to \frac{(6.02\times {10}^{23})}{(12+33)\times 1000}200$ molecules

$\Rightarrow (\frac{NA200}{44\times 1000}-{10}^{21})$ molecules of $C{O}_2$

$\frac{1}{NA}(\frac{Na200}{44\times 1000}-{10}^{21})$ of moles $C{o}_2$

$\Rightarrow \frac{1}{6.02\times {10}^{23}}(\frac{6.02\times {10}^{23}\times 2}{44\times 10}-{10}^{23})$ of moles of $C{O}_2$

$\Rightarrow (\frac{2}{44\times 10}-\frac{1}{60.2}\times {10}^2)$ of moles of $C{O}_2?$

$\Rightarrow (\frac{1}{220}-\frac{1}{602})$ moles of $C{O}_2$

$\Rightarrow 0.00288425$ moles of $C{O}_2$

$\Rightarrow 2.88{/}^{-3}moles$ of ${10}_2$

$\Rightarrow n=2.88$