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Question

Chemistry Question on Some basic concepts of chemistry

If 224 mL of a triatomic gas has a mass of 1 g at 273 K and 1 atm. pressure, then the mass of one atom is

A

8.30×0−23g8.30 \times 0^{-23}g

B

6.24×10−23g6.24 \times 10^{-23}g

C

2.08×10−23g2.08 \times 10^{-23}g

D

5.54×10−23g5.54 \times 10^{-23}g

Answer

5.54×10−23g5.54 \times 10^{-23}g

Explanation

Solution

No. of mol of triatomic gas =224 mL22400 mL mol−1=\frac{224\,mL}{22400\, mL\, mol^{-1} } =10−2 mol=10^{-2}\,mol No. of molecules =10−2 mol×(6.02×1023=10^{-2} \, mol \times (6.02\times 10^{23} molecules mol−1)mol^{-1}) =6.02×1021=6.02 \times 10^{21} molecules No. of atoms =3×6.02×1021=18.06×1021=3 \times 6.02\times10^{21}=18.06\times10^{21} =1.806×1021=1.806\times 10^{21} 1.806×10211.806\times10^{21} atoms has mass =1 g=1\,g 11 atom has mass =11.806×1021g=\frac{1}{1.806 \times 10^{21}} g =10×10−221.806g=0.554×10−22g=\frac{10 \times 10^{-22}{1.806}}g=0.554\times 10^{-22}g =5.54×10−23=5.54\times 10^{-23}