Explanation

(c)
$pV = nRT$or $\ \ \ \ \ \ \ \ \ \ \ \ \ pV = \frac{w}{M}RT$
or $M = \frac{w}{V}\frac{RT}{p}$
or $M = d\frac{RT}{p}$
$d = 1.964\ g/dm^{3} = 1.964 \times 10^{- 3}\ g/cc$
$p = 76\ cm\ Hg = 1\ atm$
$R = 0.0812\ L\ atm\ K^{- 1}\ {mol}^{- 1}$
$= 82.1\ cc\ atm\ K^{- 1}\ {mol}^{- 1}$
$T = 273\ K$
$M = \frac{1.964 \times 10^{- 3} \times 82.1 \times 273}{1} = 44$
The molecular weight of ${CO}_{2}$ is 44.
So, the gas is ${CO}_{2}.$