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After reading a number of explanations about it after googling, I'm a bit confused on this. Some explanations are doing it using a solution density, others aren't, though most are. On one site it's showing different densities for a solution of a compound at different w/w% values. I'm now very lost as to how you convert from a w/w% to molarity. Can someone clarify?

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Weight percent is essentially the same as (I arbitrarily picked grams as the mass unit):

$$ \%~\mathrm{w/w~X} = \mathrm{g~X\over g~solution} $$

In order to get to the $\mathrm{mol\over L}$ units of molarity, you have to convert by multiplying by the solution density, $\rho$, and dividing by the molecular weight of $\mathrm X$, $M_\mathrm{X}$:

$$ \mathrm{g~X\over g~solution} \times \stackrel{\left(1\over M_\mathrm X\right)}{\mathrm{mol~X\over g~X}} \times \stackrel{\left(\rho\right)}{\mathrm{g~solution\over L~solution}} = \mathrm{mol~X\over L~solution} $$

One key thing to remember is that $\rho$ is the density of the solution, not of the pure solvent. If you're working with dilute solutions (below $\sim\!100~\mathrm{mM}$ for aqueous) these two densities will be similar, but in concentrated solutions they will diverge.

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  • $\begingroup$ I'd rather write mass fraction $w(\ce{X}) = m(\ce{X})/m(\text{solution})$, which is obviously not in percent. I also would call it (amount) concentration, $c(\ce{X}) = n(\ce{X})/V(\text{solution})$ for clarity. Then you can also easily see that the density of the solution is $\rho(\text{solution}) = m(\text{solution})/V(\text{solution})$. Your dimensional analysis works fine though. (In your first equation you need to multiply the right hand side with $100\%$, otherwise you're using mass fraction.) $\endgroup$ – Martin - マーチン Nov 8 '17 at 4:55
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  1. Use the density of the solution to get the total volume of the solution.
  2. Then use the weight percent of solute to determine the amount of substance of the solute.
  3. Use the amount of substance of the solute divided by the volume to get molarity.

Molarity is relevant to our concept of chemistry and kinetics of reactions. However wt/wt% is frequently easier to use on an industrial scale.

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W/W% is just the same as a any percentage but now the difference is that let say that a concentrated solution contains 67% w/w ammonia that means that there is 67g of ammonia in 100g of the solution; now let us consider a question

A concentrated solution of aqueous ammonia is $28.0\%\text{w/w }\ce{NH3}$ and has a density of $\pu{0.899 g/mL}$. What is the amount concentration of $\ce{NH3}$ in this solution?

Converting $28.0\%\text{w/w}$ to ratio form gives us $28/100$ then you multiply it with the density $\pu{0.899 g/ml}$ for ammonia and $\pu{17.04 g}$ is an equivalent of one mole.

$$28/100 \times \frac{\pu{0.899 g}}{\pu{1 ml}} \times \frac{\pu{1 mol}}{\pu{17.04 g}} \times \frac{\pu{1000 ml}}{\pu{1ml}}$$

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