Question

Calculate the moles and the mass of solute in each of the following solutions. (a) \(250.0 \mathrm{~mL}\) of \(1.50 \mathrm{M} \mathrm{KCl}\) (b) \(250.0 \mathrm{~mL}\) of \(2.05 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\)

Short answer

(a) The moles of solute in \( 250.0 \mathrm{~mL} \) of \( 1.50 \mathrm{M} \mathrm{KCl} \) are \( 0.375 \mathrm{moles} \) and the mass of the solute is \( 27.95 \mathrm{~g}. \) \n(b) The moles of solute in \( 250.0 \mathrm{~mL} \) of \( 2.05 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4} \) are \( 0.5125 \mathrm{moles} \) and the mass of the solute is \( 72.79 \mathrm{~g}. \)

Step-by-step solution

Calculate the moles of solute for both solutions

(a) The moles of \( \mathrm{KCl} \) can be calculated by multiplying the volume (converted into liters) by the molarity: Moles = \(0.250 \mathrm{L} \times 1.50 \mathrm{M} = 0.375 \mathrm{moles} \) \n (b) The moles of \( \mathrm{Na}_{2} \mathrm{SO}_{4} \) can be calculated similarly: Moles = \(0.250 \mathrm{L} \times 2.05 \mathrm{M} = 0.5125 \mathrm{moles} \)

Calculate the molar mass of solutes

The molar mass of \( \mathrm{KCl} \) is \( 39.10 \mathrm{~g/mol} \) for K and \( 35.45 \mathrm{~g/mol} \) for Cl, giving a total of \( 74.55 \mathrm{~g/mol} \). The molar mass of \( \mathrm{Na}_{2}\mathrm{SO}_{4} \) is determined by adding the weights of 2 Na (\( 2 \times 22.99 \mathrm{~g/mol} \)), 4 O (\( 4 \times 16.00 \mathrm{~g/mol} \)) and 1 S \(\left(32.07 \mathrm{~g/mol}\right)\), which equates to \( 142.04 \mathrm{~g/mol}\).

Calculate the mass of the solute for both solutions

(a) The mass of \( \mathrm{KCl} \) can be obtained by multiplying moles of solute by the molar mass: Mass = \( 0.375 \mathrm{moles} \times 74.55 \mathrm{g/mol} = 27.95 \mathrm{g} \) \n (b) The mass of \( \mathrm{Na}_{2}\mathrm{SO}_{4} \) can also be calculated similarly: Mass = \( 0.5125 \mathrm{moles} \times 142.04 \mathrm{g/mol} = 72.79 \mathrm{g} \)

Key concepts

Calculating Moles

Understanding the mole concept is fundamental in chemistry, as it allows us to measure amounts of substances in a useful way. A mole is like a counting unit for chemistry, similar to a dozen for eggs. In chemistry, a mole refers to Avogadro's number, which is approximately 6.022 x 10^23 particles of a substance.

When calculating moles, we often use the molarity of a solution (moles of solute per liter of solution, M). For example, if we have a 1.50M KCl solution, we calculate the moles of KCl in a given volume by multiplying the volume by the molarity, just as done in our exercise.

Step-by-Step Calculation

Let's say we have 250.0 mL of a 1.50 M KCl solution. First, convert the volume from milliliters to liters (250.0 mL is 0.250 liters), as molarity is defined per liter of solution. Now, simply multiply the volume in liters by the molarity: Moles of KCl = 0.250 L x 1.50 M = 0.375 moles. This straightforward approach applies to any solute in any molar solution.

Molar Mass

The molar mass is another essential concept and represents the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It can be determined by summing the atomic masses of all atoms in a molecule.

In the provided exercise, we calculate the molar mass of KCl by adding the atomic masses of potassium (K) and chlorine (Cl), which are 39.10 g/mol and 35.45 g/mol respectively. Thus, the molar mass of KCl is 74.55 g/mol.

Finding Molar Mass

For a more complex compound like Na2SO4, we perform the same addition of atomic masses: 2 x 22.99 g/mol for sodium (Na), 4 x 16.00 g/mol for oxygen (O), and 32.07 g/mol for sulfur (S), totaling 142.04 g/mol. Knowing the molar mass allows us to convert between grams and moles, which is crucial for stoichiometric calculations and preparing solutions.

Solution Concentration

Solution concentration describes how much solute is present in a given amount of solvent. Molarity (M), one of the most common concentration measures, is the number of moles of solute per liter of solution.

Concentration allows us to understand the strength of a solution. A higher molarity means more solute particles in the same amount of solvent, which usually results in a more reactive or potent solution.

Practical Application

In practice, you might prepare a solution by dissolving a known mass of solute to achieve a desired molarity. Conversely, if given the molarity, as in our exercise (e.g., a 2.05M Na2SO4 solution), you can determine the amount of solute in a specific volume, thereby enabling accurate predictions about the solution's behavior in chemical reactions.