Mass of salt required to make a 1.2 g/ml solution
To answer in detail Jan, I didn't make stranged calculus, now I answered more than a year ago I can't tell you if my calculus are the same, but here is the logic.



For any solution (aqueous) we have,
$$m_{solution}=\rho_{solution}\times V \tag1$$

We also have,
$$\rho_{solution}=d\times \rho_{water} \tag2$$

where d is the relative density of the solution and then, merging both $(1)$ and $(2)$,
$$m_{solution}=d \times \rho_{water}\times V \tag3$$

The mass of solute in the solution is a percentage of the masse of the whole solution then, because,

$$n_{solute}=\frac{m_{solute}}{M_{solute}} \tag4$$

We have,

$$n_{solute}=\frac{p\times d \times \rho_{water}\times V}{100\times M_{solute}}\tag5$$

Then find my previous result with $C\times V=n_{solute}$. 



Now I think everyone is able to goes from molar to mass concentration then I can stop here. If not ask me.