Question

When \(V \mathrm{ml}\) of \(2.2 \mathrm{M}-\mathrm{H}_{2} \mathrm{SO}_{4}\) solution is mixed with \(10 \mathrm{~V} \mathrm{ml}\) of water, the volume contraction of \(2 \%\) takes place. The molarity of diluted solution is (a) \(0.2 \mathrm{M}\) (b) \(0.204 \mathrm{M}\) (c) \(0.196 \mathrm{M}\) (d) \(0.224 \mathrm{M}\)

Short answer

The molarity of the diluted solution is 0.204 M, so the correct answer is (b) 0.204 M.

Step-by-step solution

- Understand the Concept of Molarity and Volume Contraction

Molarity is defined as the number of moles of solute per liter of solution. When two solutions are mixed, their volumes are not always additive due to volume contraction. Volume contraction refers to the phenomenon where the combined volume of two mixed liquids is less than the sum of their individual volumes. In this case, a 2% volume contraction occurs when the sulfuric acid solution is mixed with water.

- Calculate the Final Volume Considering Volume Contraction

The initial total volume before contraction would be the sum of the volume of the acid, which is \(V\) ml, and the volume of water, which is \(10V\) ml. Hence, total initial volume is \(V + 10V = 11V\) ml. Since there is a 2% volume contraction, the final volume after contraction will be 98% of the initial total volume. The final volume \(V_f\) is thus given by: \[V_f = 98\% \times 11V = 0.98 \times 11V\]

- Calculate the Moles of Sulfuric Acid in the Original Solution

To find the molarity after dilution, we first need to know how many moles of solute (sulfuric acid, \(H_2SO_4\)) are in the original solution. Since the original concentration is 2.2 M, the number of moles of \(H_2SO_4\) in \(V\) ml is given by the equation: \[\text{moles of } H_2SO_4 = Molarity \times Volume \; \text{in liters} = 2.2 \times \frac{V}{1000}\]

- Calculate the Molarity of the Diluted Solution

The molarity after dilution can be calculated using the moles of solute (from step 3) and the final volume after contraction (from step 2). The molarity \(M_f\) is given by: \[M_f = \frac{\text{moles of } H_2SO_4}{V_f\;\text{in liters}} = \frac{2.2 \times V / 1000}{0.98 \times 11V / 1000} = \frac{2.2}{0.98 \times 11}\]

- Simplify to Find the Molarity

Now simplify the equation for \(M_f\) to find the molarity of the diluted solution: \[M_f = \frac{2.2}{0.98 \times 11} = \frac{2.2}{10.78} = 0.204 \;M\] Therefore, the molarity of the diluted solution is 0.204 M.

Key concepts

Understanding Volume Contraction

The concept of volume contraction is a bit like when you mix sand with stones; the total volume isn't just all the sand plus all the stones because the sand fills in the gaps between the stones. In chemistry, when we mix liquids like water and sulfuric acid (\( \text{H}_2\text{SO}_4 \)), their molecules interact in a way that they can occupy less space than the sum of their separate volumes. This phenomenon is called volume contraction. It's important when calculating the final volume of a solution after mixing, as we need to account for this slight decrease. Imagine you have a container with a solution, and after mixing it with another, the level doesn't go as high as you expected. That's volume contraction at work. In our problem, knowing that the total volume contracts by 2% after mixing is crucial for accurately determining the molarity of the final solution.

In most classroom settings, volume contraction might not be heavily emphasized, but it is quite relevant in industrial applications where precise volumes of solutions are required.

The Process of Solution Dilution

Think of solution dilution as making your favorite concentrate juice less intense by adding water. We dilute solutions to decrease the concentration of solutes—in our case, sulfuric acid. By adding more solvent (which is usually water), we spread the solute molecules further apart, lowering their concentration per unit volume. The concept of solution dilution is applied to adjust the molarity, or moles of solute per liter, to a desired level. The process is fundamental in various domains, from basic laboratory work to manufacturing and even culinary arts. It's key to maintain the balance between the volume of solute and the total volume of the newly diluted solution. For our example, adding water to the acidic solution reduces its molarity, but we have to account for volume contraction to get the final, precise molarity value. Solution dilution is not as simple as 'just add water'—precision is essential.

A common misunderstanding in dilution problems is neglecting the change in volume after mixing—always remember to factor in any volume changes!

Moles of Solute in Solution

Now, to grasp the moles of solute, imagine you have a bag containing exactly twelve apples; that's like having a mole of apples. In chemistry, a mole is a specific quantity—about 6.02 x 10^23 particles—of a substance. For solutions, we speak of moles of solute, which are the dissolved particles within the solvent. Determining the moles of solute is vital for understanding solution concentration. It's how chemists quantify substances rather than just using 'parts' or 'pieces'—it's a way for us to compare apples to apples, or in our case, molecules to molecules. In the textbook problem, the moles of sulfuric acid before dilution are calculated using the volume of the solution and its initial molarity. The central idea is that the amount of solute—in moles—stays constant even when you dilute the solution; only its dispersion within the solvent changes.

Remember that the volume used in molarity calculations should be in liters, so converting milliliters to liters is a common step. By learning how to calculate moles of solute, students gain insight into the quantitative analysis of solutions, opening doors to comprehending more complex chemical reactions and preparations.