C O _ 2 gas is liquefied at 27 ^ o C . Gas follows the following...

$\left(A\right)$ Maximum density of $C{O}_{2}$ in gaseous phase will be at point 'A'
Now: $d = \dfrac{PM}{RT} = \dfrac{60 \times 44}{R \times 300} = 107gm / lt = 0.107gm / cc$
Or $d = \dfrac{massofgas}{Volumeofgas} = \dfrac{44gm}{440cc} = 0.1$
$\left(B\right)$ Density of liquid $C{O}_{2}$ at 60 atm (at point B)
At point B there will be only liquid $C{O}_{2}$ . So we can find out the density of density of liquid $C{O}_{2}$ only at point B.
$\text{d} = \dfrac{M}{V} = \dfrac{44}{0.044} = 1\text{gm/cc}$
$\left(C\right)$ at point C the volume will be $\dfrac{0.44 + 0.044}{2} \approx 0.22\text{L}$
And pressure is same so obviously approx. half of $C{O}_{2}$ is liquified
$\left(D\right)$ Z can be greater than 1 at $2{7}^{ο}C$. For example at point 'A' Z is 1.07, So wrong.