More Examples on Molarity

Calculating Molar Concentrations from the Mass of Solute

Distilled white vinegar (see image below) is a solution of acetic acid, CH₃CO₂H, in water. A 0.500-L vinegar solution contains 25.2 g of acetic acid. What is the concentration of the acetic acid solution in units of molarity?

Distilled white vinegar is a solution of acetic acid in water. Image credit: OpenStax, Chemistry

Solution

As in previous examples, the definition of molarity is the primary equation used to calculate the quantity sought. In this case, the mass of solute is provided instead of its molar amount, so we must use the solute’s molar mass to obtain the amount of solute in moles:

\(M = \cfrac{\text{mol solute}}{\text{L solution}} \)\(= \cfrac{25.2\text{ g }\mathrm{CH_3CO_2H} × \frac{1\text{ mol }\mathrm{CH_3CO_2H}}{60.052\text{ g }\mathrm{CH_3CO_2H}}}{0.500\text{ L solution}} \)\(= 0.839 M\)

\(M = \cfrac{\text{mol solute}}{\text{L solution}} = 0.839 M\)

\(M = \cfrac{0.839\text{ mol solute}}{1.00\text{ L solution}}\)

Determining the Mass of Solute in a Given Volume of Solution

How many grams of NaCl are contained in 0.250 L of a 5.30-M solution?

Solution

The volume and molarity of the solution are specified, so the amount (mol) of solute is easily computed as demonstrated in the example on deriving moles and volumes from molar concentrations from the previous lesson.

\(M = \cfrac{\text{mol solute}}{\text{L solution}}\)

\(\text{mol solute} = M × \text{L solution}\)

\(\text{mol solute} = 5.30 \frac{\text{mol NaCl}}{\text{L}} × 0.250\text{ L} \)\(= 1.325\text{ mol NaCl}\)

Finally, this molar amount is used to derive the mass of NaCl:

\(1.325\text{ mol NaCl} × \cfrac{58.44\text{ g NaCl}}{\text{mol NaCl}} \)\(= 77.4\text{ g NaCl}\)

When performing calculations stepwise, as in the example above, it is important to refrain from rounding any intermediate calculation results, which can lead to rounding errors in the final result. In the example above, the molar amount of NaCl computed in the first step, 1.325 mol, would be properly rounded to 1.32 mol if it were to be reported. However, although the last digit (5) is not significant, it must be retained as a guard digit in the intermediate calculation. If we had not retained this guard digit, the final calculation for the mass of NaCl would have been 77.1 g, a difference of 0.3 g.

In addition to retaining a guard digit for intermediate calculations, we can also avoid rounding errors by performing computations in a single step (see the example below). This eliminates intermediate steps so that only the final result is rounded.

Determining the Volume of Solution Containing a Given Mass of Solute

In first example, we found the typical concentration of vinegar to be 0.839 M. What volume of vinegar contains 75.6 g of acetic acid?

Solution

First, use the molar mass to calculate moles of acetic acid from the given mass:

\(\text{g solute} × \cfrac{\text{mol solute}}{\text{g solute}} = \text{mol solute}\)

Then, use the molarity of the solution to calculate the volume of solution containing this molar amount of solute:

\(\text{mol solute} × \cfrac{\text{L solution}}{\text{mol solute}} = \text{L solution}\)

Combining these two steps into one yields:

\(\text{g solute} × \cfrac{\text{mol solute}}{\text{g solute}} × \cfrac{\text{L solution}}{\text{mol solute}} \)\(= \text{L solution}\)

\(75.6\text{ g }\mathrm{CH_3CO_2H} \left ( \cfrac{\text{mol }\mathrm{CH_3CO_2H}}{60.05\text{ g}} \right ) \left ( \cfrac{\text{L solution}}{0.839\text{ mol }\mathrm{CH_3CO_2H}} \right ) \)\(= 1.50\text{ L solution}\)