Question

$112.0\, mL$ of $NO_2$ at $STP$ was liquefied, the density of the liquid being $1.15\, g \,mL^{-1}$. Calculate the volume and the number of molecules in the liquid $NO_2.$

• A. $0.10\, mL$ and $3.01 × 10^{22}$
• B. $0.20\, mL$ and $3.01 × 10^{21}$
• C. $0.20\, mL$ and $6.02 × 10^{23}$
• D. $0.10\, mL$ and $6.02 × 10^{21}$

Answer 1

Answer: (B)

At $STP \,22400\, mL$ of $NO_2 ≡ 46 \,g$ of $NO_2$
$\therefore 112.0\,mL\,of\,NO_{2}=\frac{112.0\,mL\times46.0\,g}{22400\,mL}=0.23\,g.$
$V_{NO_2}=\frac{mass}{density}=\frac{0.23}{1.15\,g\,mL^{-1}}=0.20\,mL.$
Number of molecules $=\frac{0.23}{46}\times6.02\times10^{23}=3.01\times10^{21}$