# How does the molarity of a solution change on boiling off the solvent?

The dilution equation $$M_1V_2=M_2V_2$$ is an equation that can be used to calculate how adding water can impact the volume and molarity of a solution. So I was wondering if this equation would apply if you were to remove the amount of solvent through means such as boiling.

For instance, suppose I have a $\pu{100 mL}$ solution of $\ce{HCl}$ that has water as its solvent and has a molarity of $\pu{3.0 M}$. Would the molarity become $\pu{6.0 M}$ if were to boil out $\pu{50 mL}$ of water of the solvent, or would it have a different $M$ value?

• That's a bad example. By boiling this solution, you will remove not just water, but also HCl. – Ivan Neretin Apr 6 '18 at 7:07
• It is probably better to refer to a better form of your equation, i.e. $$c = \frac{n}{V} \implies n = c\cdot V.$$ Where $c$ is the concentration (or if in $\pu{mol/L}$ molarity; better not to use $M$ for this as it commonly refers to the molecular mass in $\pu{g/mol}$), $n$ is the amount of substance, $V$ is the volume. Keeping $n$ constant and varying in $c$ and $V$ will therefore give you $$c_1\cdot V_1 = c_2\cdot V_2.$$ The way in which you change the volume is (in principle) irrelevant. (It should be noted, that the solvent and solute should endure this kind of change.) – Martin - マーチン Apr 6 '18 at 9:45