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.
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 $1000~\pu{ml}$.
