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Is there a way of converting mass percentage of a solution to obtain its density? (With mass percentage been given and the compounds which are mixed to form the binary solution)

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    $\begingroup$ No.$\mathstrut$ $\endgroup$ – Ivan Neretin Nov 13 '17 at 11:17
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    $\begingroup$ Related: chemistry.stackexchange.com/q/50269 $\endgroup$ – Jan Nov 13 '17 at 12:23
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    $\begingroup$ @IvanNeretin: Huh? There are plenty of cases where an accurate solution density can be calculated from the relevant densities and mass percentage of solute. Plenty. $\endgroup$ – Gert Nov 13 '17 at 16:31
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If we assume no significant change in volume when mixing, then the answer is yes.

Assume we have $w$ percent of a solute (suffixed as $1$), dissolved in a solvent (suffixed as $2)$. That means that $100\ \mathrm{g}$ of solution contains $w\ \mathrm{g}$ of solute and $(100-w)\ \mathrm{g}$ of solvent.

The density $d$ of the solution (for $100\ \mathrm{g}$ of solution) is calculated as:

$$d=\frac{m}{V}=\frac{100}{V_1+V_2}$$

With the respective densities $d_1$ and $d_2$, $V_1=\frac{w}{d_1}$ and $V_2=\frac{100-w}{d_2}$:

$$d=\frac{100}{\frac{w}{d_1}+\frac{100-w}{d_2}}$$

Slightly reworked:

$$d=\frac{100d_1d_2}{wd_2+(100-w)d_1}=\frac{100d_1d_2}{w(d_2-d_1)+100d_1}$$

As stated, this approximation works only if there's no significant change in volume during mixing. It works very well for example for sucrose in water solution. Should it turn out that the mixed volume $V \neq V_1+V_2$, an inaccuracy/error will arise.

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There is no simple rule for that, but a linear interpolation. This interpolation is quite a good model for the components without significant intermolecular interactions, like nonpolar solvent mix. Polar solvents and/or compounds often manifest strong interactions and solvent contractions, related to nonzero mixing enthalpy. E.g mixing of 1 L of ethanol and 1L of water produce ( after dissipation of mixing heat ) not 2L, but about 1.9 L.

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