Mass of salt required to make a 1.2 g/ml solution I would like to make an NaCl and water solution with a density of 1.2 g/ml. How would you calculate the grams of NaCl required to make a 1000 ml solution?

I have used density of mixture = (mass NaCl + mass H2O)/1000 ml and solved for the mass of NaCl. However, the resulting mass does not give a density of 1.2 g/ml in practice (and in fact does not dissolve) so is there another way of calculating this? (at room temperature (~21 °C))
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.