Problem 36 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

A bottle of commercial sulphuric acid (density \(1.787 \mathrm{~g} / \mathrm{mL}\) ) is labelled as 86 percent by weight. What is the molarity of the acid? What volume of the acid has to be used to make 1 litre of \(0.2 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4} ?\)

Step-by-Step Solution

Answer

The molarity of the acid is 15.67 M, and 12.76 mL of it is needed.

Step 1: Determine the Mass of 1 L of the Acid

We begin by assuming that we have 1 liter of the sulphuric acid solution. Given that the density is \(1.787 \ \mathrm{g/mL} \), we can find the mass of 1 liter (which is 1000 mL) by multiplying the volume by the density. The calculation is as follows: \[mass = 1000 \ \mathrm{mL} \times 1.787 \ \mathrm{g/mL} = 1787 \ \mathrm{g}\]

Step 2: Calculate the Mass of Pure Sulphuric Acid

From the label, we know the solution is 86% sulphuric acid by weight. To find the mass of pure \( \mathrm{H_2SO_4} \) in 1 liter, we take 86% of the total mass: \[mass \ of \ \mathrm{H_2SO_4} = 1787 \ \mathrm{g} \times 0.86 = 1536.82 \ \mathrm{g}\]

Step 3: Calculate the Moles of Sulphuric Acid

The next step is to convert the mass of \( \mathrm{H_2SO_4} \) to moles. The molar mass of \( \mathrm{H_2SO_4} \) is approximately 98.08 \( \mathrm{g/mol} \). Calculate the moles using:\[moles \ of \ \mathrm{H_2SO_4} = \frac{1536.82 \ \mathrm{g}}{98.08 \ \mathrm{g/mol}} \approx 15.67 \ \mathrm{mol}\]

Step 4: Determine the Molarity

Molarity \( (M) \) is defined as moles of solute per liter of solution. Since we have the number of moles in 1 liter of the solution, the molarity is:\[Molarity \ (\mathrm{M}) = \frac{15.67 \ \mathrm{mol}}{1 \ \mathrm{L}} = 15.67 \ \mathrm{M}\]

Step 5: Use the Dilution Formula

To make 1 L of a \(0.2 \ \mathrm{M} \) solution, use the dilution formula: \( M_1V_1 = M_2V_2 \), where \( M_1 = 15.67 \ \mathrm{M} \), \( M_2 = 0.2 \ \mathrm{M} \), and \( V_2 = 1 \ \mathrm{L} \). Solve for \( V_1 \), the volume of the concentrated acid needed:\[15.67 \mathrm{M} \times V_1 = 0.2 \mathrm{M} \times 1 \mathrm{L}\]Solve for \( V_1 \):\[V_1 = \frac{0.2 \ \mathrm{L}}{15.67} \approx 0.01276 \ \mathrm{L} = 12.76 \ \mathrm{mL}\]

Step 6: Verify the Final Molarity

It is good practice to verify the calculations. First ensure that the desired molarity from the dilution calculation is matched:Substitute \( V_1 = 12.76 \ \mathrm{mL} \) into the dilution equation to ensure consistency:\[15.67 \mathrm{M} \times 0.01276 \ \mathrm{L} = 0.2 \mathrm{M} \times 1 \mathrm{L}\]Both sides should yield the same value, confirming correctness.

Key Concepts

Sulphuric Acid

Sulphuric acid is a highly important and widely used chemical in various industries. It is known for its strong acidic properties and is used in everything from manufacturing fertilizers to refining petroleum. The chemical formula for sulphuric acid is \( \mathrm{H_2SO_4} \), indicating that each molecule consists of two hydrogen atoms, one sulfur atom, and four oxygen atoms. This compound is a vital component in many chemical reactions.
Everyday applications of sulphuric acid highlight its versatile nature:

• Fertilizer Production: It's a key ingredient in the production of phosphate fertilizers.
• Petroleum Refining: Used in processing crude oil to remove impurities.
• Metal Processing: Involved in cleaning metals and in manufacturing batteries.
• Domestic Products: Found in drain cleaners and has roles in pH regulation of swimming pools.

However, sulphuric acid must be handled with care. It's highly corrosive to skin and metals. Always use safety gear when working with it to prevent burns or other injuries.
The strong acidic nature means it can participate in many hasty reactions, making it essential in laboratory settings. Given its high utility and demand, it is often produced in large quantities globally.

Density and Concentration

Understanding density and concentration is vital in chemistry, especially when working with solutions like sulphuric acid. Density refers to the mass of a substance divided by its volume and is expressed as \( \mathrm{g/mL} \) or \( \mathrm{kg/m^3} \). This property helps determine how much of a substance is present in a given volume. In the case of the given sulphuric acid, its density is 1.787 \( \mathrm{g/mL} \). This means every milliliter of the acid weighs 1.787 grams.
Concentration, on the other hand, describes the amount of a substance (solute) dissolved in a unit volume of solution. It's often expressed in terms of molarity (\( \mathrm{M} \)), where 1 molarity implies one mole of solute per liter of solution. The exercise involves calculating the molarity from a given percentage by weight and density.

• Step 1: Calculate the mass of the solution using density.
• Step 2: Use the percentage by weight to find the mass of pure sulphuric acid in the solution.
• Step 3: Convert this mass into moles using the molar mass of sulphuric acid.
• Step 4: Finally, calculate the molarity as moles per liter of the solution.

This layered process offers a clear path from raw data to concentration, bridging the gap between density and molarity with precision.

Dilution Formula

The dilution formula is a simple yet powerful tool in chemistry used to create solutions of a desired molarity or concentration from a more concentrated stock solution. The formula, \( M_1V_1 = M_2V_2 \), is rooted in the idea that the amount of solute remains constant before and after dilution.
Here's how each component works:

• \( M_1 \): Initial molarity of the concentrated solution.
• \( V_1 \): Volume of the concentrated solution needed.
• \( M_2 \): Desired molarity of the final diluted solution.
• \( V_2 \): Total volume of the final solution.

In practice, to prepare a more dilute solution, you determine the volume of concentrated solution needed by rearranging the formula to solve for \( V_1 \):\[V_1 = \frac{M_2 \times V_2}{M_1}\]
An example from the exercise involved creating a 0.2M sulphuric acid solution from the concentrated form. Given that \( M_1 \) was 15.67M, and \( M_2 \) was 0.2M with \( V_2 \) as 1 L, it was essential to calculate \( V_1 \). The answer, 12.76 mL, showed how little of the concentrated acid was needed.
By efficiently using this formula, chemists can prepare solutions without wasting resources, maintaining the necessary balance of careful measurement and precision essential in mastering chemical formulations.