A) 28 g
B) \[14\times 22.4\text{ }g\]
C) 56 g
D) None of these
Correct Answer: A
Solution :
From ideal gas equation, we have \[PV=nRT\] where P is pressure, V the volume, R the gas constant, T the temperature and n the number of moles. \[\therefore \] \[n=\frac{PV}{RT}\] Given,\[P=22.4\]atm pressure \[=22.4\times 1.01\times {{10}^{5}}N/{{m}^{2}}\] \[V=2L=2\times {{10}^{-3}}{{m}^{3}},\] \[R=8.31J/mol-K\] \[T=273\text{ }K\] \[\therefore \] \[n=\frac{22.4\times 1.01\times {{10}^{5}}\times 2\times {{10}^{-3}}}{8.31\times 273}\] \[n=1.99\approx 2\] Since, \[n=\frac{mass}{atomic\text{ }weight}\] we have \[mass=n\times atomic\text{ }weight=2\times 14=28\text{ }g\]
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