Molar mass and molar volume of monochloroethylene
You don't need to know the volume to determine the molar volume $V_\mathrm{m}$ of a substance. In your notations:

$$V_\mathrm{m} = \frac{M}{d} = \frac{\pu{62.37 g mol-1}}{\pu{2.56 g L-1}} = \pu{24.36 L mol-1}$$

Alternatively, you could've just used ideal gas law:

$$V_\mathrm{m} = \frac{V}{n} = \frac{RT}{P} = \frac{\pu{0.0821 atm L mol-1 K-1}\cdot\pu{295.8 K}}{\pu{0.9968 atm}} = \pu{24.36 L mol-1}$$

P. S. You determined molecular mass correctly, but, as MaxW pointed out, the units are wrong: you probably made a typo and put $\pu{g L-1}$ instead of $\pu{g mol-1}$.