Problem 22 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

\(29.2 \%(\mathrm{w} / \mathrm{w}) \mathrm{HCl}\) stock solution has a density of \(1.25 \mathrm{~g} \mathrm{~mL}^{-1}\). The molecular weight of \(\mathrm{HCl}\) is \(36.5 \mathrm{~g} \mathrm{~mol}^{-1}\). The volume \((\mathrm{mL})\) of stock solution required to prepare a \(200 \mathrm{~mL}\) solution of \(0.4 \mathrm{M} \mathrm{HCl}\) is :

Step-by-Step Solution

Answer

8 mL of stock solution is needed.

Step 1: Determine the mass of HCl in the stock solution

The stock solution has a concentration of \(29.2\%\) by weight, meaning \(29.2\)-grams of HCl are present per \(100\)-grams of solution. Given the density \(1.25 \, \text{g/ml}\), find out how much HCl is present in \(1\, \text{liter} (1000\, \text{mL})\) of this solution. First, calculate the mass of \(1\, \text{liter}\) of the solution as follows:\[ \text{Mass of 1L stock solution} = 1000\, \text{mL} \times 1.25\, \text{g/mL} = 1250\, \text{g} \]Next, calculate the mass of HCl in \(1\, \text{liter}\):\[ \text{Mass of HCl in 1L} = 1250\, \text{g} \times \frac{29.2}{100} = 365\, \text{g} \]

Step 2: Calculate the molarity of the stock solution

Using the mass of HCl in \(1\, \text{liter}\) from Step 1, calculate the molarity:\[ \text{Molarity} = \frac{\text{mass of HCl in 1L}}{\text{molecular weight}} = \frac{365\, \text{g}}{36.5\, \text{g/mol}} = 10\, \text{mol/L} \]

Step 3: Use dilution formula to find required volume of stock solution

Use the dilution equation: \( M_1V_1 = M_2V_2 \), where:- \(M_1 = 10\, \text{M}\) (molarity of stock solution)- \(V_1 = \text{required volume of stock solution} (mL)\)- \(M_2 = 0.4\, \text{M}\) (desired molarity)- \(V_2 = 200\, \text{mL}\) (desired volume)Plug in the values:\[ 10 \times V_1 = 0.4 \times 200 \]Solve for \(V_1\):\[ V_1 = \frac{0.4 \times 200}{10} = 8\, \text{mL} \]

Key Concepts

Molarity

Molarity is a term that describes the concentration of a solution. It is defined as the number of moles of solute per liter of solution. In simple terms, molarity tells you how many molecules of a particular substance are present in a certain volume of liquid. This concept is crucial in chemistry, especially when preparing solutions required for experiments or reactions.
To calculate the molarity (\(M\)), you need two pieces of information:

• The amount of solute in moles
• The volume of the solution in liters

The formula to determine molarity is:\[M = \frac{\text{moles of solute}}{\text{liters of solution}}\]In the original exercise, we calculated the molarity of HCl in a stock solution. We converted the mass of HCl into moles by using its molecular weight, and then used this value to find the molarity by considering the total volume of the solution. Knowing molarity helps us predict how the solution will react under different conditions.

Density

Density is a property that describes how much mass is contained in a given volume of a substance. It is an essential concept because it allows us to relate the mass of a substance to its volume, which is particularly helpful in laboratory settings.
In basic terms, density (\(\rho\)) is calculated using the formula:\[\rho = \frac{\text{mass}}{\text{volume}}\]The original exercise involved calculating how much mass is present in a given volume of hydrochloric acid solution. With the given density of \(1.25 \text{ g/mL}\), we were able to find out the mass of the solution for any specific volume. For instance, to find the mass in one liter (or 1000 mL), we multiplied the volume by the density, yielding \(1250 \text{ g}\) as the total mass. This calculation was vital for further determining the amount of HCl present in the whole solution.

Dilution

Dilution is the process of decreasing the concentration of a solute in a solution, usually by adding more solvent. This is a common practice in laboratories when a particular concentration is required for an experiment or reaction.
The principle behind dilution is that the amount of solute remains the same before and after the process. Therefore, you can use the dilution equation:\[M_1V_1 = M_2V_2\]Where:

• \(M_1\) is the molarity of the initial (more concentrated) solution
• \(V_1\) is the volume of the initial solution
• \(M_2\) is the molarity of the diluted (less concentrated) solution
• \(V_2\) is the volume of the diluted solution

In the original exercise, this formula was used to find \(V_1\), the volume of the stock solution needed to achieve a desired concentration in a new solution. By plugging in known values, we calculated that 8 mL of stock solution was needed to prepare a 200 mL solution with a molarity of \(0.4 \text{ M}\). This calculation is critical in ensuring the accurate preparation of chemical solutions with precise concentrations.