$\left(\mathrm{A}\right)$ Maximum density of $\mathrm{C}{\mathrm{O}}_2$ in gaseous phase will be at point 'A'
Now: $\mathrm{d}=\frac{\mathrm{P}\mathrm{M}}{\mathrm{R}\mathrm{T}}=\frac{60\times 44}{\mathrm{R}\times 300}=107\mathrm{ }\mathrm{g}\mathrm{m}/\mathrm{l}\mathrm{t}=0.107\mathrm{ }\mathrm{g}\mathrm{m}/\mathrm{c}\mathrm{c}$
Or $\mathrm{d}=\frac{\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{g}\mathrm{a}\mathrm{s}}{\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{g}\mathrm{a}\mathrm{s}}=\frac{44\mathrm{ }\mathrm{g}\mathrm{m}}{440\mathrm{ }\mathrm{c}\mathrm{c}}=0.1$
$\left(\mathrm{B}\right)$ Density of liquid $\mathrm{C}{\mathrm{O}}_2$ at 60 atm (at point B)
At point B there will be only liquid $\mathrm{C}{\mathrm{O}}_2$ . So we can find out the density of density of liquid $\mathrm{C}{\mathrm{O}}_2$ only at point B.
$\text{d}=\frac{\text{M}}{\text{V}}=\frac{44}{0.044}=1\text{ gm/cc}$
$\left(\mathrm{C}\right)$ at point C the volume will be $\frac{0.44+0.044}{2}\approx 0.22\text{L}$
And pressure is same so obviously approx. half of $\mathrm{C}{\mathrm{O}}_2$ is liquified
$\left(\mathrm{D}\right)$ Z can be greater than 1 at $27^{\mathrm{o}}\mathrm{C}$. For example at point 'A' Z is 1.07, So wrong.