Question

What volume of \(0.120 \mathrm{M} \mathrm{CuSO}_{4}\) is required to give \(0.150 \mathrm{~mol}\) of copper(II) sulfate, \(\mathrm{CuSO}_{4} ?\)

Short answer

1.25 L of the 0.120 M CuSO₄ solution is needed.

Step-by-step solution

Identify the relationship between moles and volume

The relationship between moles \((n)\) of a solute, the molarity \((M)\) of a solution, and the volume \((V)\) of the solution in liters is given by the formula: \[ n = M \times V \]. We will use this equation to find the required volume of the \(0.120 \: \mathrm{M} \: \mathrm{CuSO}_{4}\) solution.

Rearrange the formula to solve for volume

We need to find the volume \((V)\), so rearrange the formula: \[ V = \frac{n}{M} \].

Substitute the known values

Substitute the known values into the equation: \(n = 0.150 \: \mathrm{mol}\) and \(M = 0.120 \: \mathrm{M}\). Thus, \[ V = \frac{0.150}{0.120} \].

Calculate the volume

Carry out the division to find the volume: \[ V = \frac{0.150}{0.120} = 1.25 \: \, \mathrm{L} \].

Key concepts

Moles

In chemistry, a "mole" refers to a specific number of particles, whether they are atoms, molecules, or ions. It is equivalent to Avogadro's number, which is approximately \(6.022 \times 10^{23}\) particles. Understanding moles is crucial when calculating chemical reactions because it bridges the gap between the atomic world and the macroscopic amounts we can measure. - Annoyed with counting atoms? The mole lets chemists 'count' particles in a simple, manageable way!- It's similar to how a dozen is 12. A mole is \(6.022 \times 10^{23}\)!When you know the number of moles, you have a count of particles that can be used to determine weights and reactions. In the given problem, we want \(0.150\) moles of copper(II) sulfate. That's \(0.150 \times 6.022 \times 10^{23}\) molecules of \(\text{CuSO}_4\)! Thinking in moles simplifies complex calculations, helping us better visualize reactions.

Volume

Volume refers to the amount of space that a substance or object occupies. In the context of solutions, it's usually measured in liters (L) or milliliters (mL). - When we mix substances, volume helps us keep track of concentrations.- In labs, we often use graduated cylinders or pipettes for precise volume measurements.The problem requires finding the volume of \(0.120 \, \text{M} \, \text{CuSO}_4\) solution needed to provide \(0.150\) moles of copper(II) sulfate. By rearranging the molarity equation, we solve for volume using \[V = \frac{n}{M}\].In this equation:- \(n\) = number of moles (0.150 moles)- \(M\) = molarity (0.120 M)By calculating, we find that the volume needed is \(1.25\; \text{L}\). Being comfortable with these volume calculations deepens your understanding of solution preparations.

Concentration calculation

Concentration describes the amount of solute present in a solution relative to the total volume of that solution. It's key in determining how "strong" or "dilute" a solution is. Molarity (\(M\)) is a common measure of concentration used in chemistry.- Molarity formula: \(M = \frac{n}{V}\)- Here, \(n\) is moles of solute, \(V\) is volume in liters.- Knowing molarity helps assess how many molecules are involved in reactions.For the exercise problem:- We know \(n = 0.150\; \text{mol}\) of \(\text{CuSO}_4\)- We want \(0.150\; \text{mol}\) in a \(0.120\; \text{M}\) solution- By rearranging the formula as \(V = \frac{n}{M}\), we find \(V = \frac{0.150}{0.120} = 1.25\; \text{L}\).Concentration calculations help chemists mix solutions accurately to achieve the desired properties and reactions.