Question

Calculate the molarity of each of the following solutions: (a) \(29.0 \mathrm{~g}\) of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) in \(545 \mathrm{~mL}\) of solution, (b) \(15.4 \mathrm{~g}\) of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) in \(74.0 \mathrm{~mL}\) of solution, (c) \(9.00 \mathrm{~g}\) of sodium chloride \((\mathrm{NaCl})\) in \(86.4 \mathrm{~mL}\) of solution.

Short answer

(a) The molarity of the ethanol solution is 1.16 M. \n(b) The molarity of the sucrose solution is 0.61 M. \n(c) The molarity of the sodium chloride solution is 1.78 M.

Step-by-step solution

Converting mass to moles

First, the mass of the solute needs to be converted into moles. One can accomplish this by dividing the mass of the solute by its molar mass: \n(a) The molar mass of ethanol (C2H5OH) is \(46.07 g/mol\). Therefore, there are \(29.0 g \div 46.07 g/mol = 0.63 moles\) of ethanol. \n(b) The molar mass of sucrose (C12H22O11) is \(342.3 g/mol\). Therefore, there are \(15.4 g \div 342.3 g/mol = 0.045 moles\) of sucrose. \n(c) The molar mass of sodium chloride (NaCl) is \(58.44 g/mol\). Therefore, there are \(9.00 g \div 58.44 g/mol = 0.154 moles\) of sodium chloride.

Converting volume to liters

Next, the volume of the solution in milliliters should be converted into liters by dividing by 1000: \n(a) \(545 mL = 0.545 liters\) \n(b) \(74.0 mL = 0.074 liters\) \n(c) \(86.4 mL = 0.0864 liters\)

Calculating Molarity

Lastly, divide the number of moles of the solute by the volume of the solution in liters. \n(a) Molarity of ethanol = \(0.63 moles \div 0.545 liters = 1.16 M\)\n(b) Molarity of sucrose = \(0.045 moles \div 0.074 liters = 0.61 M\)\n(c) Molarity of sodium chloride = \(0.154 moles \div 0.0864 liters = 1.78 M\)

Key concepts

Molar Mass

Understanding molar mass is fundamental to solving problems in chemistry, especially when calculating the composition of a solution. Simply put, the molar mass is the weight of one mole of a substance, usually expressed in grams per mole (\text{g/mol}).

For instance, in ethanol (\text{C}_2\text{H}_5\text{OH}), carbon (\text{C}) has an atomic mass of about 12.01 \text{g/mol}, hydrogen (\text{H}) is about 1.008 \text{g/mol}, and oxygen (\text{O}) is approximately 16.00 \text{g/mol}. Therefore, a single molecule of ethanol, which has two carbons, six hydrogens, and one oxygen, will have a total molar mass of \(2\times12.01 + 6\times1.008 + 16.00\) g/mol, adding up to 46.07 \text{g/mol}.

Knowing the molar mass enables us to connect the microscopic concept of atoms and molecules to macroscopic quantities of matter that we can measure.

Moles Conversion

The concept of moles is a bridge between the atomic scale and the real world. A mole represents Avogadro's number (\text{6.022} \times \text{10}^{23}) of particles, be they atoms, ions, or molecules. Converting grams to moles or vice versa is essential for quantitative chemistry analysis.

The formula to convert mass (in grams) to moles is:
\[ \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
For example, to convert 29.0 grams of ethanol to moles, divide the mass by its molar mass, resulting in:\[ 29.0 \text{g} \div 46.07 \text{g/mol} = 0.63 \text{moles} \]
This is the first step in many stoichiometry problems, where the actual amount of substances involved in reactions needs to be quantified.

Solution Concentration

Solution concentration, particularly molarity, is one of the most widely used terms in chemistry for describing the strength of a solution. Molarity (M) specifies the number of moles of a solute present in one liter of solution, expressed as moles per liter (\text{mol/L}).

It's computed using the formula:
\[ \text{Molarity (M)} = \frac{\text{number of moles of solute}}{\text{volume of solution in liters}} \]
For example, to find the molarity of ethanol in the given problem, divide the moles of ethanol (0.63) by the volume of the solution in liters (0.545 L), yielding a molarity of 1.16 M.

Understanding molarity is vital as it allows chemists to predict how a substance will react in a solution, which is crucial for experiments and practical applications in industries.

Chemistry Stoichiometry

Chemistry stoichiometry is all about the quantitative relationships between the amounts of reactants and products in a chemical reaction. It hinges on the law of conservation of mass and the concept of moles.

Stoichiometry allows us to predict the outcome of chemical reactions, figure out limiting reactants, and calculate percent yields. It takes into account the molar mass of reactants and products and requires a balanced chemical equation to relate moles of different substances. It's a tool that chemists use to quantify reactions and is indispensable in laboratory and industrial settings for preparing compounds in the correct proportions.

Working through exercises such as the calculation of molarity from given masses and volumes trains students in practical stoichiometry, reinforcing the abstract concepts with tangible practice.