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

Answer: (D)

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