Problem 33 - Chemistry The Science in Context

Question

Water is allowed to evaporate from \(100.0 \mathrm{mL}\) of \(0.24 M\) \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) until the solution volume is \(60.0 \mathrm{mL} .\) What is the molar concentration of the evaporated solution?

Step-by-Step Solution

Answer

Answer: The molar concentration of the evaporated solution is 0.40 M.

Step 1: Calculate the initial amount of solute in the solution

We know the initial volume (100 mL) and molar concentration (0.24 M) of the solution. We can calculate the moles of Na2SO4 present in the initial solution using the formula, M = n/V, where M is the molar concentration, n is the moles of solute, and V is the volume of the solution in liters. Rearrange the equation to solve for n: n = M * V. n = 0.24 M * 100.0 mL * (1 L/1000 mL) n = 0.024 moles of Na2SO4

Step 2: Determine the final volume of the solution

After evaporation, the volume of the solution is given as 60.0 mL. We need to convert this volume into liters for our calculations. Final volume = 60.0 mL * (1 L/1000 mL) Final volume = 0.060 L

Step 3: Calculate the molar concentration of the evaporated solution

Since the amount of solute (0.024 moles Na2SO4) remains constant, we can calculate the final molar concentration by using the formula M = n/V with the final volume. M_final = (0.024 moles Na2SO4) / (0.060 L) M_final = 0.40 M So, the molar concentration of the evaporated solution is 0.40 M.

Key Concepts

Solution Concentration

Understanding the concentration of a solution is fundamental in chemistry as it tells us how much solute is present in a given volume of solvent. The concentration of a substance is often expressed in molarity, which is the number of moles of solute per liter of solution, represented as M. One mole of a substance is equivalent to its molecular weight in grams, which is a constant value specific to each chemical compound.

To find the molarity, we use the equation:
\( M = \frac{n}{V} \)
where \( M \) stands for molarity, \( n \) for the number of moles of solute, and \( V \) the volume of the solution in liters. This equation is a crucial formula in solution chemistry and helps us understand the relationship between the solute and the solution's volume. For real-world applications, such as preparing a saline solution for medical use or diluting a substance for a chemical reaction, calculating the molarity accurately is vital.

Evaporation Effect on Concentration

When a solvent, like water, evaporates from a solution, the volume of the solution decreases, but the amount of solute remains unchanged. This affects the solute concentration—specifically, the molarity of the solution. Since molarity is dependent on the volume of the solution (as seen in the formula \( M = \frac{n}{V} \)), a decrease in volume due to evaporation will result in an increase in the molarity.This phenomenon is because the moles of solute are now dispersed in a smaller volume, making the solution more concentrated. It is important to distinguish that evaporation does not alter the number of moles of solute; it merely concentrates those moles in less solvent. This concept is crucial in processes like concentration of fruit juices, where removing water enhances flavors, or in chemical synthesis, where achieving certain molarity is necessary for the reaction to occur correctly.

Calculating Moles of Solute

Moles of solute in a solution can be calculated using the equation \( n = M \times V \), where \( n \) represents the number of moles of solute, \( M \) the molarity of the solution, and \( V \) the volume of the solution in liters. Calculating moles is an essential step in determining concentration and participating in stoichiometric calculations in chemical reactions.

In practice, this means that if you know the molarity of your solution and the volume of your solvent, you can easily compute the amount of solute present. For example, a 0.24 M solution of sodium sulfate (Na2SO4) in 100.0 mL of water contains 0.024 moles of sodium sulfate. This calculation is straightforward but plays an integral role in making precise solutions for laboratory experiments, industrial processes, or pharmacological formulations.