Explanation

(a) From ideal gas equation, we have
$pV = nRT$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n = \frac{pV}{RT}$
Given, $p = 22.4\ atm\ pressure$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 22.4 \times 1.01 \times 10^{5}\ Nm^{- 2}$,
$V = 2L = 2 \times 10^{- 3}\ m^{3}$,
$R = 8.31\ J\ mol^{- 1} - K^{- 1}$,
$T = 273$ 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, $n = \ \frac{Mass}{Atomic\ weight}$
We have,
$mass = n \times atomic\ weight = 2 \times 14 = 28$ g