Mixing two solutions: how to solve such problems, is there a trick? [closed]

Two containers were filled with salt solutions, and the first container contained 1 liter less solution than the second. The mass concentration of salt from the solution in the first container was 10%, and in the second, 20%. After both solutions were poured into the third container, a new solution was obtained, the concentration of which was 16%. How much solution (volume) was in each container initially?

I just wonder how such kind of problems are solved. I'd be very grateful if you help me to understand. It's from 7th grade.

• This is mostly a linear equations math problem. Let $x$ be volume of 10% solution and $y$ be the volume of the 20% solution. Now you need to figure out two equations that relate $x$ and $y$. // The only "chemistry" part is assuming that the volumes of the different solutions are additive, which isn't absolutely true. However the overriding consideration is that the concentrations of the salt solutions have only been given to two significant figures, so assuming the volumes are additive is reasonable.
– MaxW
Commented Sep 27, 2023 at 21:55
• @MaxW But volume additivity is not the only problem, as the mass percentage refers to the solution mass, not volume, so solution density would be needed. It is possible the task neglects density difference. Commented Sep 28, 2023 at 3:18
• For 2 kg 10% solution is needed 3 kg of 20% solution. What particular masses to use to have 1 L difference depends on densities of both solutions. 2 and 3 L would be applicable only if densities of 10% and 20% solutions are the same, what is not true. But it is possible the task is oversimplified for the 7th grade and they go this way. Commented Sep 28, 2023 at 3:33
• @Poutnik - You're correct, nice catch! There isn't enough information given in the problem for a solution. The problem is that the solution concentrations are expressed as weight %'s, but the difference in the two solutions is given in liters. In order to make the problem solvable the difference between the solutions should have been given as a mass, not a volume.
– MaxW
Commented Sep 28, 2023 at 7:05

@MaxW showed the procedure for doing this problem in the comments, so I'll just show that procedure applied to your specific example. We will assume the volumes are additive.

let $$a$$ represent the volume in liters of the first solution, $$b$$ represent the volume in liters of the second solution, and $$c$$ represent the volume in liters of the combined solution.

We have two equations:

$$.1a + .2b = .16c$$

$$a+b=c$$

We know from the problem that $$a=b-1$$, so let's make that substitution:

$$.1(b-1)+.2b=.16c$$

$$b-1+b=c$$

Now, lets solve that bottom equation for c, so we can write everything in terms of b:

$$2b-1=c$$

Now, substituting $$c=2b-1$$ to the first equation:

$$.1(b-1)+.2b=.16(2b-1)$$

Now, we just solve for b:

$$.1b-.1+.2b=.32b-.16$$

$$.3b-.1=.32b-.16$$

$$.06=.02b$$

$$b=3$$

Now that we have b, we can easily solve for a:

$$a=b-1=3-1=2$$

So, there were intially 2 liters of the first solution and 3 liters of the second solution.

The task description is flawed, as it incorrectly assumes 10% or 20% solutions contain $$\pu{100 g/L}$$ resp. $$\pu{200 g/L}$$ and that mixing of solutions does not change the total volume.

But if they used

• either mass percentages and solution masses
• or mass concentrations and assumption $$ΔV=0$$

then they could directly use the mixing rule:

$$\frac{m_\mathrm{10\%}}{m_\mathrm{20\%}} = \frac{|20\%-16\%|}{|10\%-16\%|} = \frac{2}{3}$$

$$\frac{V_\pu{100 g L-1}}{V_\pu{200 g L-1}} = \frac{|\pu{200 g L-1} - \pu{160 g L-1}|}{|\pu{100 g L-1} - \pu{160 g L-1}|} = \frac{2}{3}$$

and directly used as the task solution:

• Mass percentage:
• 10% solution: $$\pu{2 kg}$$
• 20% solution: $$\pu{3 kg}$$.
• Mass concentration:
• $$\pu{100 g L-1}$$ solution: $$\pu{2 L}$$
• $$\pu{200 g L-1}$$ solution: $$\pu{3 L}$$.

For the mass percentage case, solution densities would be needed to calculate volumes.