Question

From 115 g of a $N{O}_2\left(g\right)$ sample, $1.2044\times {10}^{24}$ molecules of $N{O}_2\left(g\right)$ are removed. Find the volume of the left over $N{O}_2\left(g\right)$ at STP.

• A. 11.2 L
• B. 22.4 L
• C. 44.8 L
• D. 61.2 L

Answer 1

The correct option is A 11.2 L

The number of moles of $N{O}_2\left(g\right)$ present in 115 g of $N{O}_2\left(g\right)$,
$n=\frac{mass}{molarmass}=\frac{115g}{46g/mol}=2.5mol$

The number of moles of $N{O}_2\left(g\right)$ removed,
$n=\frac{givennumberofmolecules}{N_A}=\frac{1.2044\times {10}^{24}}{6.022\times {10}^{23}}=2mole$

The number of moles of $N{O}_2\left(g\right)$ left,
= 2.5 - 2 = 0.5 mol

Volume of $N{O}_2\left(g\right)$ left at STP.
$V=n\times 22.4L=11.2L$