Composition of Substances and Solutions

Molarity and Dilutions

An aqueous solution is composed of a material dissolved in water. This type of solution can be described in terms of molarity.

In practical chemistry, it is often useful to dissolve compounds to make solutions. To dissolve a substance means to incorporate it into a liquid so as to form a solution. The substance in which another substance dissolves is the solvent. The dissolved material in a solution is the solute. An aqueous solution is a solution in which the solvent is water. Aqueous solutions are the most common type of solution because water dissolves a large variety of solutes.

Varying amounts of a solute may be dissolved in a given volume of water. The amount of solute dissolved in a given volume of solvent is its concentration. Concentration can be measured in many ways, but the most common measurement of concentration in chemistry is molarity (M), which is the number of moles of a solute dissolved in 1 liter of water. For example, 1 mole of sodium chloride (NaCl) dissolved in 1 liter of water has a molarity of 1 mol/L, or 1 molar (M). The concentration of any solution can be calculated by dividing the total number of moles of solute by the total volume.
molarity(M)=number of moles of solutevolume of solution\text{molarity}\;(\rm{M})=\frac{\text{number of moles of solute}}{\text{volume of solution}}
Step-By-Step Example
Calculate the Molarity of an Aqueous Solution of Sodium Chloride
What is the molarity of a solution made by dissolving 3.2 g of NaCl in enough water to produce 225 mL of solution?
Step 1
Convert the given mass of NaCl to moles, using the periodic table to determine the molar mass of NaCl, 58.44 g/mol.
3.2gNaCl×1molNaCl58.44gNaCl=0.00548molNaCl3.2\;\rm{g}\;\rm{ NaCl}\times\frac{1\;\rm{mol}\;\rm{NaCl}}{58.44\;\rm g\;\rm{NaCl}}=0.00548\;\rm {mol}\;\rm{NaCl}
Step 2
Convert the given milliliters of solution to liters.
(225mL solution)(1L1000mL)=0.225L solution(225\;\rm{mL}\text{ solution})\left(\frac{1\,\rm{ L}}{1000\,\rm{ mL}}\right)=0.225\;\rm{L}\text{ solution}
Step 3
Divide the number of moles of solute by the volume of solution, in liters, to find the molarity.
molarity=0.0548molNaCl0.225L=0.24MNaCl\text{molarity}=\frac{0.0548\;\rm{mol}\;\rm{NaCl}}{0.225\;\rm{L}}=0.24\;\rm M\;\rm{NaCl}
The molarity of the solution is 0.24 M NaCl.

For any aqueous solution, there is a maximum amount of solute that can be dissolved in the solution. A saturated solution contains the maximum amount of dissolved solute normally possible at a certain temperature. An unsaturated solution contains less than the maximum amount of dissolved solute normally possible at a certain temperature. Notice that saturation depends on the temperature of the solution. Heating a solution will allow more solute to be dissolved, even if the solution had been saturated at a lower temperature. A supersaturated solution contains an amount of dissolved solute greater than is normally possible at a certain temperature.

Any aqueous solution can be made less concentrated by adding water. To dilute a solution is to decrease its concentration by adding more solvent to it. A less concentrated solution made by diluting a solution is a dilution. Dilutions can be described using a formula in which the initial product of molarity and volume equals the final product of molarity and volume.
M represents molarity, V represents volume, and the numbers 1 and 2 indicate the initial and final values respectively.
Step-By-Step Example
Calculate the Concentration of a Dilution of Sodium Chloride in Water
Consider a 1 M solution of NaCl in 1 liter of water. Another 1 liter of water is added to the solution. What is the concentration of the dilution?
Step 1
Substitute the given values into the dilution equation. Let x represent the unknown molarity.
(1M)(1L)=(xM)(2L)(1\;\rm M)(1\;\rm L)=(\it x\,\rm M)(2\;\rm L)
Step 2
Divide both sides of the equation by 2 L to solve for x.
xM=(1M)(1L)(2L)=0.5MNaCl\begin{aligned}\rm{\it{x}\rm{M}}&=\frac{(1\;\rm M)(1\;\rm L)}{(2\;\rm L)}\\&=0.5\;\rm M\;\rm{NaCl}\end{aligned}
The concentration of the dilution is 0.5 M NaCl.
An aqueous solution can be made more concentrated by allowing the water to evaporate. To let the solvent evaporate from a solution to increase its concentration is to concentrate it.
Step-By-Step Example
Calculate the Concentration of a Sodium Chloride Solution
Suppose 1 liter of a 1.5 M solution of NaCl is evaporated to half a liter. What is the concentration of the remaining solution?
Step 1
Substitute the given values into the dilution equation.
(1.5M)(1L)=(xM)(0.5L)(1.5\;\rm M)(1\;\rm L)=\it(x\,\rm M)(0.5\;\rm L)
Step 2
Let xx represent the unknown molarity, and solve the equation for xx.
xM=(1.5M)(1L)0.5L=3MNaCl\begin{aligned}x\,\rm M&=\frac{(1.5\;\rm M)(1\;\rm L)}{0.5\;\rm L}\\&=3\;\rm M\;\rm{NaCl}\end{aligned}
The concentration of the remaining solution is 3 M NaCl. In this case, the concentration of the solution can be calculated without knowing the identity of the solvent.

If the water evaporates completely, crystallization of the solute can occur. Crystallization is the formation of a solid in which the particles form a highly organized structure. Crystallization can also occur in a supersaturated solution as the solution cools. As conditions become unfavorable for the solute to remain in solution, small crystals begin to form. In some cases, these crystals rapidly expand.

Most of the time, it is impossible to count the moles of a solute to calculate the molarity of the solution. Instead, scientists use the molar mass of the solute to determine the concentration.

Step-By-Step Example
Calculate the Amount of Solute Needed to Produce a Solution of Desired Quantity and Molarity
The molar mass of NaCl is 58.44 g/mol. How much solute is needed to make 5.80 liters of a 2.00 M solution?
Step 1
Use the molarity to convert liters of solution to moles of solution.
(5.80L)(2.00mol1L)=11.6mol(5.80\;\rm{L})\left(\frac{2.00\;\rm{mol}}{1\;\rm L}\right)=11.6\;\rm{mol}
Step 2
Multiply by the molar mass to find the mass of NaCl.
(11.6mol)(58.44gNaCl1mol)=678gNaCl(11.6\;\rm{mol})\left(\frac{58.44\;\rm g\;\rm{NaCl}}{1\;\rm{mol}}\right)=\;678\;\rm g\;\rm{NaCl}
To make the desired solution, 678 g NaCl is needed.