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](https://chemistry.stackexchange.com/questions/47625/mass-of-salt-required-to-make-a-1-2-g-ml-solution?noredirect=1#comment148460_47625), 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.




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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](https://en.wikipedia.org/wiki/Relative_density) 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 $C\times V=n\_{solute}$. 




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


 3

The answer is: as much as water can dissolve.


The density of saturated sodium chloride in water at 21°C is $1.20~\pu{g cm^{-3}}$.


So you may take roughly $1000~\pu{ml}$ water of 21°C and put in sodium chloride such that not all of it can be dissolved (it will take some 400 grams). From the resulting solution‘s supernatant take $1000~\pu{ml}$.


 3