Question

Calculate the molarity of each solution. a. 3.25 mol of LiCl in 2.78 L solution b. 28.33 \(\mathrm{gC}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) in 1.28 \(\mathrm{L}\) of solution c. 32.4 \(\mathrm{mg} \mathrm{NaCl}\) in 122.4 \(\mathrm{mL}\) of solution

Short answer

a. Molarity of LiCl solution = 1.17 M, b. Molarity of glucose solution = 0.125 M, c. Molarity of NaCl solution = 4.54 mM

Step-by-step solution

Calculate the molarity of LiCl solution

To find the molarity of a solution, use the formula \[\text{Molarity (M)} = \frac{\text{Amount of solute in moles}}{\text{Volume of solution in liters}}\]. For LiCl, the amount of solute is 3.25 moles and the volume is 2.78 L, so \[M_{\text{LiCl}} = \frac{3.25 \text{ mol}}{2.78 \text{ L}}\].

Calculate the molarity of glucose solution

First convert the mass of solute to moles using its molar mass (Glucose \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) has a molar mass of 180.16 g/mol). Then divide by the volume in liters. \[\text{Moles of glucose} = \frac{28.33 \text{ g}}{180.16 \text{ g/mol}}\], then compute molarity \[M_{\text{glucose}} = \frac{\text{moles of glucose}}{1.28 \text{ L}}\].

Calculate the molarity of NaCl solution

Convert the mass from milligrams to grams, then to moles (NaCl has a molar mass of 58.44 g/mol). Convert the volume from mL to L. Calculate the molarity: \[\text{Moles of NaCl} = \frac{32.4 \text{ mg} \times \frac{1 \text{ g}}{1000 \text{ mg}}}{58.44 \text{ g/mol}}\], then \[M_{\text{NaCl}} = \frac{\text{moles of NaCl}}{0.1224 \text{ L}}\].

Key concepts

Solution Concentration

When discussing solution concentration, we are referring to the amount of a solute that is present in a given volume of solvent. This is a fundamental concept in chemistry, as it directly impacts the solution's properties and potential reactions. One common way to express concentration is molarity, which is the number of moles of solute per liter of solution.

To calculate molarity, you divide the moles of solute by the volume of the solution in liters. It is important to pay attention to units during calculations to ensure accuracy. In the case of 3.25 mol of LiCl in 2.78 L of solution, for instance, the molarity would be calculated using the formula:
\[ \text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} = \frac{3.25 \text{ mol}}{2.78 \text{ L}}\].

Understanding how to calculate molarity allows students to predict how a given concentration will behave in chemical reactions, which is critical for experiments and industrial applications alike.

Molecular Weight

Molecular weight, also known as molecular mass, is the sum of the atomic weights of all atoms in a molecule. Each element's atomic weight can be found on the periodic table. This value is crucial for converting between moles of a substance and the corresponding mass in grams, which is a common task in chemical problem-solving.

For example, to convert 28.33 grams of glucose (\(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)) into moles, identify its molecular weight first. The molecular weight of glucose is calculated by adding up the atomic weights of carbon (C), hydrogen (H), and oxygen (O), resulting in 180.16 g/mol. Understanding the relationship between mass and moles via molecular weight is essential for accurate molarity calculations.

Mole-to-Gram Conversion

To convert moles of a substance to grams, you multiply the number of moles by the substance's molecular weight. This is expressed by the formula: \[\text{Mass in grams} = \text{Moles} \times \text{Molecular Weight}\]. Conversely, to find the number of moles when the mass is given, you divide the mass by the molecular weight.

For instance, converting the mass of glucose to moles involves dividing by its molecular weight:
\[\text{Moles of glucose} = \frac{28.33 \text{ g}}{180.16 \text{ g/mol}}\]. It is vital to perform this conversion to determine molarity, as evidenced in the original problem with glucose and NaCl solutions.

Dilution of Solutions

Diluting a solution involves adding more solvent to the solution, which consequently decreases its concentration. This process is important to understand because it allows for the creation of a solution with a desired concentration from a more concentrated starting point.

The concept of dilution can also be employed when adjusting the concentrations of the solutions in laboratory settings. The relationship between the concentration and volume before and after dilution is given by the formula: \[ M_{1}V_{1} = M_{2}V_{2} \],
where \( M_{1} \) and \( V_{1} \) are the molarity and volume before dilution, and \( M_{2} \) and \( V_{2} \) are the molarity and volume after dilution. Mastery of this formula is necessary for accurately preparing solutions of known concentrations.