# How to calculate the total concentration after mixing two solutions of differing dilutions?

If you mix $$\pu{20 ml}$$ of $$\pu{3 M}$$ sugar solution with $$\pu{30 ml}$$ of a $$\pu{5 M}$$ sugar solution, what solution do you end up with?

What I did,

\begin{align} \text{volume} &= \frac{\text{amount of substance}}{\text{concentration}}\\ \dfrac{3}{0.02} &= 150\\ \dfrac{5}{0.03} &= 166.\overline{6}\\ 150+166.\overline{6} &= 316.\overline{6} \end{align}

This is the wrong answer. Any formulas which could help would be appreciated.

• Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms". (cc @safdar ) – Martin - マーチン Aug 8 '20 at 17:10
• Please also note that descriptive terms or names of quantities shall not be arranged in the form of an equation; i.e. do not write "$\text{volume}=\frac{\text{amount of substance}}{\text{concentration}}$"; write $V=\frac nc$ instead. – user7951 Aug 8 '20 at 17:43

Remember two solutions of different concentrations are mixed together, the total amount of substance of the solution is the sum of the amounts of the individual solutions. when you get the sum of the amounts, you add their volumes and use those two to determine the new concentration of the solution. $$n_1=0.02\ \mathrm l\times3\ \mathrm{mol\ l^{-1}}=0.06\ \mathrm{mol}$$ $$n_2=0.03\ \mathrm l\times5\ \mathrm{mol\ l^{-1}}=0.15\ \mathrm{mol}$$ $$n_\text{total}=n_1+n_2=0.15\ \mathrm{mol}+0.06\ \mathrm{mol}=0.21\ \mathrm{mol}$$ New concentration: $$c_\text{new}=\frac{n_\text{total}}{V_\text{total}}=\frac{0.21\ \mathrm{mol}}{0.05\ \mathrm l}= 4.2\ \mathrm{mol\ l^{-1}}$$