# How to Solve Chemistry Solutions Problems?

Do you often have difficulties when you have to solve chemistry solutions problems or concentration problems, perhaps to pass from one concentration expressed as a mass / mass percentage to another, expressed in moles per liter of solution? Do you see how molarity is smoke in the eyes? No fear.

Follow me and with a little patience, you will see that after all it is not so difficult to find your way. I will propose problems divided by type; with the respective methodology proposed for the solution.

Case 1: Passing from a concentrated molar solution to a diluted one, that is, “dilution”.

Example:

35 mL of a 2.0 M sulfuric acid solution is taken and placed in a 500 mL volumetric flask. Calculate the molarity of the solution thus obtained.

To solve this type of exercise, you have to understand an important concept: the number of moles of solute taken are equal to the number of moles of the solute present in the new solution. What changes after the operation is the concentration, which becomes smaller (if it is expressed in the same unit of measurement).

Remembering that the molar concentration is obtained by dividing the number of moles of solute by the volume of solution (Important! expressed in liters or better in dm3 ):

M = n / V

We have that n = M × V

In the first solution described by the problem, the 2 molar one, the number of moles of the acid will be:

n initials = initial M × initial V

In the diluted solution, you will have:

n final = M final × V final

We understood, however, that the moles do not change, so we will have that:

n initial = n final

So we can write:

M initial × V initial = M final × V final

By examining the data of the problem, we see that our unknown is the final molarity. Therefore, the solution of the problem will be given by the expression:

Final M = initial M × initial V / final V

Pay attention to two important Points.

First Point: The expression above is an expression and not a proportion:

Second Point: the volume, before carrying out the calculations, must be transformed into liters or cubic decimeters.

Now mathematically,

Final M = 2 mol × dm -3 × 0.035 dm 3 / 0.500 dm 3 = 0.14 mol × dm -3 i.e. 0.14 M.

Case 2: Pass from a concentrated solution, expressed in percent mass by mass to a dilute one, expressed in moles per liter

Example:

From a solution of HCl at 37% m / m, density 1.19 g / mL, 50 mL are taken, placed in a volumetric flask and brought to a final volume of 250 mL. Calculate the molarity of the solution thus obtained.

This is a bit more “tough” than previous one, but do not worry.

What do the data tell us about the problem? “Let’s read them” carefully. Density is the mass of the unit of volume and is calculated:

d = m / V

Density relates the volume of the solution taken to its mass. To say that the density is 1.19 g / mL is as saying that one-milliliter of this solution has a mass of 1.19 grams.

Since we have taken 50 mL, the product of the density and the volume will give the corresponding mass:

m = d × V = 1.19 g × mL -1 × 50 mL = 59.5 g

This mass represents the mass of the sample, i.e. of the solution (solute + solvent): it must therefore be remembered that the solution is a homogeneous mixture, that is, it is not made up of a pure liquid. The data of the problem tells us that only 37% of the mass of this solution is represented by hydrochloric acid, that is:

Mass of acid = 59.5 g × 37/100 = 22 g

Since the problem asks us to express the final concentration in molarity, we transform the amount of acid into moles:

n HCl = m HCl / molar mass HCl = 22 g / 36.46 g × mol -1 = 0.60 mol

One last effort: the final volume of the solution is given to us in milliliters; we must therefore transform it into liters:

250 mL = 0.250 L

The molarity of the final solution will be:

M = n / V = ​​0.60 mol / 0.250 L = 2.4 M

Case 3: Calculate the molarity of a solution whose concentration is expressed as a mass / mass percent.

Example

Calculate the molarity of a 22.38% nitric acid solution having a density of 1.130 g / mL.

It is not difficult: since we have to calculate the molarity, the volume of which is expressed in liters or cubic decimeters, for our convenience we take into consideration, as an arbitrary volume, one liter.

One liter of solution will have a mass of 1130 grams, i.e. 1.130 g / mL × 1000 mL (i.e. 1 L).

Now, as in type 2, we calculate the mass of solute that is 22.38% of the total mass of the solution:

m HNO3 = 1130 g × 22.38 / 100 = 252.9 g

To know the molar concentration, just transform 252.9 g of nitric acid into moles of nitric acid, remembering that this quantity of substance is contained in a liter of solution.

M = 252.9 g × L -1 / 63.01 g × mol -1 = 4.01 M