# 7.17: Concentrations: Molarity Conversions

- Page ID
- 222412

As stated previously, chemists have defined several types of concentrations, which each use a different chemically-acceptable unit, or combination of units, to indicate the amount of solute that is dissolved in a given amount of solvent. The equation that is used to calculate the molarity of a solution was presented and applied in the previous section of this chapter. In contrast to the percent-based concentrations that have been discussed, which, once calculated, are presented as "end-result" values, a molarity is a "hidden" conversion factor that can be applied in additional problem-solving contexts.

Because determining the amount of solute that is present in a homogeneous mixture is fundamental to the quantitative study of solutions, the corresponding concentration values are chemically-significant measurements. Therefore, since the molarity of a solution can be applied as a conversion factor to bring about a desired unit transformation, as will be discussed in the following paragraphs, chemists defined this type of concentration as a molar standard.

## Molarity Indicator

Recall that all molar standards have a corresponding indicator word or phrase that identifies which relationship and, therefore, which type of conversion factor, must be applied to solve the problem at-hand. As discussed in the previous section, a molarity is calculated by dividing the molar amount of solute by the volume of the solution, as a whole. During this process, no unit cancelation occurs, because the units that are present in the numerator and denominator, "mol" and "L," respectively, do not match one another. Therefore, the unit that is used to report the molarity of a solution must contain both of these units, which are related to one another in a fractional format, "mol/L." Alternatively, the unit for molarity can be abbreviated as a capital "M," which should be italicized, * M*, when typed and underlined,

__M__, if hand-written. This final symbolism is the indicator that a molarity conversion factor should be developed and applied to solve a given problem.

For example, calculate how many moles of potassium metabisulfite, K_{2}S_{2}O_{5}, are present in a 0.64 liter solution of 1.9 *M* potassium metabisulfite, K_{2}S_{2}O_{5}.

The italicized capital *"M"* indicates that a molarity conversion factor should be developed and applied to solve this problem.

## Representing Molarities as Conversion Factors

As stated above, a capital "M," which is italicized, *M*, when typed and underlined, __M__, when hand-written, is the indicator that a molarity conversion factor should be developed and applied to solve a given problem. This symbolism is the abbreviation for "mol/L," which is the unit that is used to report the molarity of a solution. The value "1.9 *M* potassium metabisulfite, K_{2}S_{2}O_{5}" can, therefore, be written as "1.9 mol/L potassium metabisulfite, K_{2}S_{2}O_{5}."

The "/" in this unit is read as "per," which is the indicator word that is associated with identifying a "hidden" conversion factor. Because the word "per" implies a ratio, which is associated with the mathematical operation of division, the word "per" is represented as a fraction bar in a conversion factor. Any number or unit read *before* the word "per" becomes the numerator of the proportion, and any number or unit found *after* the word "per" is written in the denominator. Therefore, "1.9 mol/L" can be written as

\( \dfrac{1.9 \; {\text{mol}}}{\text{L}}\) or \( \dfrac{\text{L}}{1.9 \; {\text{mol}}} \)

The conversion factor that is shown on the left is a direct representation of the indicated quantity, and the second conversion factor is derived by interchanging where each value is written, relative to the fraction bar.

Finally, recall that all of the numerical values that are incorporated into the equalities and conversion factors that are derived from molar standards must be associated with *two* units. Because a molarity is calculated by dividing the *molar amount of solute* by the *volume of the solution*, as a whole, the *chemical formula of the solute* should be incorporated as the secondary unit on the molar quantity in the corresponding conversion factors, and the word *"solution"* should be written as the secondary unit on the volumetric value in the proportions that are being developed. Since potassium metabisulfite, K_{2}S_{2}O_{5}, is the substance that is associated with the given molarity, this chemical is the solute in the solution, and, therefore, "1.9 mol/L potassium metabisulfite, K_{2}S_{2}O_{5}" can be written as

\( \dfrac{1.9 \; {\text{mol}} \; \ce{K_2S_2O_5}}{\text{L} \; \text{solution}} \) or \( \dfrac{\text{L} \; \text{solution}}{1.9 \; {\text{mol}} \; \ce{K_2S_2O_5}}\)

Again, the conversion factor that is shown on the left is a direct representation of the given molarity, and the second conversion factor is derived by interchanging where each quantity is written, relative to the fraction bar.

## Calculations

In order to calculate how many moles of potassium metabisulfite, K_{2}S_{2}O_{5}, are present in a 0.640 liter solution of 1.9 *M* potassium metabisulfite, K_{2}S_{2}O_{5}, a conversion factor based on the given molarity must be applied. Because the quantity "1.9 *M* potassium metabisulfite, K_{2}S_{2}O_{5}" should be expressed as a conversion factor, it should *not* be selected as the "given" quantity in this problem. Therefore, "0.64 liter solution" is the given quantity, and, consequently, the first molarity conversion factor that is shown above must be utilized to completely eliminate the given unit, "liter solution." Finally, recall that "hidden" conversion factors are *not *exact quantities, and therefore, the number of significant figures that they contain must be considered when applying the correct number of significant figures to a calculated quantity, as shown below.

\( {\text {0.640}}\) \({\cancel{\rm{L} \; \rm{solution}}} \times\) \( \dfrac{1.9 \; \rm{mol} \; \rm{K_2S_2O_5}}{\cancel{\rm{L} \; \rm{solution}}}\) = \( {\text {1.216}}\) \({\rm{mol} \; \rm{K_2S_2O_5}}\) ≈ \( {\text {1.2}}\) \({\rm{mol} \; \rm{K_2S_2O_5}}\)