Problem 5 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The amount of sugar \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) required to prepare \(2 \mathrm{~L}\) of its \(0.1 \mathrm{M}\) aqueous solution is: (a) \(136.8 \mathrm{~g}\) (b) \(17.1 \mathrm{~g}\) (c) \(68.4 \mathrm{~g}\) (d) \(34.2 \mathrm{~g}\)

Step-by-Step Solution

Answer

The correct answer is (c) 68.4 g.

Step 1: Understand the Molarity Formula

Molarity (M) is defined as the number of moles of solute per liter of solution. The formula is \( M = \frac{n}{V} \), where \( n \) is the number of moles of solute and \( V \) is the volume of solution in liters.

Step 2: Calculate Moles of Sugar Needed

Since we need a 0.1 M solution in 2 L, substitute the known values into the molarity formula: \( 0.1 = \frac{n}{2} \). Solving for \( n \), we get \( n = 0.1 \times 2 = 0.2 \) moles.

Step 3: Determine Molar Mass of Sugar

Calculate the molar mass of \( \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11} \). The molar masses of \( \mathrm{C} \), \( \mathrm{H} \), and \( \mathrm{O} \) are approximately 12 g/mol, 1 g/mol, and 16 g/mol, respectively. So, the molar mass of sugar is \( 12 \times 12 + 1 \times 22 + 16 \times 11 = 342 \) g/mol.

Step 4: Calculate the Mass of Sugar

Using the number of moles calculated and the molar mass, determine the mass: \( \text{Mass} = n \times \text{Molar Mass} = 0.2 \times 342 \) g. Calculating this gives \( 68.4 \) g.

Step 5: Match the Calculation with Options

Compare the calculated mass of sugar (68.4 g) with the given options: (a) 136.8 g, (b) 17.1 g, (c) 68.4 g, (d) 34.2 g. The correct answer is option (c) 68.4 g.

Key Concepts

Moles of solute

Understanding moles is crucial when dealing with solutions. A mole is simply a way to count particles, such as atoms or molecules, using Avogadro's number, which is approximately \( 6.022 \times 10^{23} \) particles per mole. In chemistry, when we talk about moles of a solute, we are referring to the amount of substance that is dissolved in a solution.
To find the moles of solute needed for a solution with a given molarity, you use the formula:

• \( \text{Molarity (M)} = \frac{\text{Moles of solute (n)}}{\text{Volume of solution (V in liters)}} \)

Applying this, if you have a molarity of 0.1 and a volume of 2 liters, you can calculate
the moles of solute as \( n = 0.1 \times 2 = 0.2 \text{ moles} \). This means you need 0.2 moles of sugar in the solution to achieve the desired concentration.

Molar mass calculation

The molar mass is key to converting between moles and grams. It tells you the mass of one mole of a given substance in grams. For sugar (\(\text{C}_{12} \text{H}_{22} \text{O}_{11}\)), you calculate it using the molar masses of carbon (C), hydrogen (H), and oxygen (O):

• Carbon: \(12 \text{ g/mol} \times 12 = 144 \text{ g/mol}\)
• Hydrogen: \(1 \text{ g/mol} \times 22 = 22 \text{ g/mol}\)
• Oxygen: \(16 \text{ g/mol} \times 11 = 176 \text{ g/mol}\)

Adding these gives the total molar mass: \( 144 + 22 + 176 = 342 \text{ g/mol}\).
With this molar mass, if you know the moles of sugar needed, you can easily convert it to grams by multiplying: \( \text{Mass} = n \times 342 \>\text{g/mol} \).

Aqueous solution preparation

Preparing an aqueous solution involves dissolving a specific amount of a compound (the solute) in water (the solvent) to achieve a desired concentration. For this problem, the goal is to prepare a 0.1 M sugar solution with a total volume of 2 liters.
Steps for preparation:

• Calculate the moles of sugar needed: This is derived from the target molarity and volume, as covered above: 0.2 moles are needed for 2 liters at 0.1 M.
• Determine the mass of sugar using its molar mass: Multiply the moles by the molar mass (342 g/mol) to find it equals 68.4 grams.
• Measure and dissolve the 68.4 grams of sugar in water: Add water until the total volume is exactly 2 liters. Stir the solution to ensure the sugar is completely dissolved.

This method ensures you have precisely the right concentration for your aqueous solution.