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Chemistry Question on Some basic concepts of chemistry

112.0 mL112.0\, mL of NO2NO_2 at STPSTP was liquefied, the density of the liquid being 1.15 g mL−11.15\, g \,mL^{-1}. Calculate the volume and the number of molecules in the liquid NO2.NO_2.

A

0.10 mL0.10\, mL and 3.01?10223.01 ? 10^{22}

B

0.20 mL0.20\, mL and 3.01?10213.01 ? 10^{21}

C

0.20 mL0.20\, mL and 6.02?10236.02 ? 10^{23}

D

0.10 mL0.10\, mL and 6.02?10216.02 ? 10^{21}

Answer

0.20 mL0.20\, mL and 3.01?10213.01 ? 10^{21}

Explanation

Solution

At STP 22400 mLSTP \,22400\, mL of NO2=46 gNO_2 = 46 \,g of NO2NO_2 ∴112.0 mL of NO2=112.0 mL×46.0 g22400 mL=0.23 g.\therefore 112.0\,mL\,of\,NO_{2}=\frac{112.0\,mL\times46.0\,g}{22400\,mL}=0.23\,g. VNO2=massdensity=0.231.15 g mL−1=0.20 mL.V_{NO_2}=\frac{mass}{density}=\frac{0.23}{1.15\,g\,mL^{-1}}=0.20\,mL. Number of molecules =0.2346×6.02×1023=3.01×1021=\frac{0.23}{46}\times6.02\times10^{23}=3.01\times10^{21}