Question

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. $8.30 \times 0^{-23}g$
• B. $6.24 \times 10^{-23}g$
• C. $2.08 \times 10^{-23}g$
• D. $5.54 \times 10^{-23}g$

Answer 1

The Correct Option is D

No. of mol of triatomic gas $=\frac{224\,mL}{22400\, mL\, mol^{-1} }$ $=10^{-2}\,mol$ No. of molecules $=10^{-2} \, mol \times (6.02\times 10^{23}$ molecules $mol^{-1})$ $=6.02 \times 10^{21}$ molecules No. of atoms $=3 \times 6.02\times10^{21}=18.06\times10^{21}$ $=1.806\times 10^{21}$ $1.806\times10^{21}$ atoms has mass $=1\,g$ $1$ atom has mass $=\frac{1}{1.806 \times 10^{21}} g$ $=\frac{10 \times 10^{-22}{1.806}}g=0.554\times 10^{-22}g$ $=5.54\times 10^{-23}$