Problem 84 - A Complete Resource Book in Chemistry for JEE Main

Question

How many grams of \(\mathrm{CH}_{3} \mathrm{OH}\) would have to be added to water to prepare \(150 \mathrm{~mL}\) of a solution that is \(2.0 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{OH} ?\) (a) \(9.6 \mathrm{~g}\) (b) \(906 \mathrm{~g}\) (c) \(4.3 \times 10^{2} \mathrm{~g}\) (d) \(9.6 \times 10^{3} \mathrm{~g}\)

Step-by-Step Solution

Answer

The answer is (a) 9.6 g.

Step 1: Understand the Molality Requirement

The problem states that we need to prepare a 2.0 M (molar) solution of \(\mathrm{CH}_{3} \mathrm{OH}\) in water. This means there should be 2.0 moles of \(\mathrm{CH}_{3} \mathrm{OH}\) per liter of solution.

Step 2: Convert Volume to Liters

The volume of the solution to be prepared is given as \(150 \mathrm{~mL}\). First, convert this volume to liters because molarity is in terms of liters: \(150 \mathrm{~mL} = 0.150 \mathrm{~L}\).

Step 3: Calculate Moles of \(\mathrm{CH}_{3} \mathrm{OH}\) Needed

Since the molarity is given as 2.0 M, use the formula for moles: \[\text{moles of } \mathrm{CH}_{3} \mathrm{OH} = \text{Molarity} \times \text{Volume in Liters} = 2.0 \mathrm{~M} \times 0.150 \mathrm{~L} = 0.300 \text{ moles}.\]

Step 4: Determine Molar Mass of \(\mathrm{CH}_{3} \mathrm{OH}\)

To convert moles to grams, determine the molar mass of \(\mathrm{CH}_{3} \mathrm{OH}\). Molar mass of \(\mathrm{CH}_{3} \mathrm{OH}\) is \(12.01\, \text{(carbon)} + 4 \times 1.01\, \text{(hydrogen)} + 16.00\, \text{(oxygen)} = 32.05 \mathrm{~g/mol}.\)

Step 5: Calculate Grams of \(\mathrm{CH}_{3} \mathrm{OH}\) Needed

Use the molar mass to find the grams of \(\mathrm{CH}_{3} \mathrm{OH}\): \[\text{Grams of } \mathrm{CH}_{3} \mathrm{OH} = \text{moles} \times \text{molar mass} = 0.300 \text{ moles} \times 32.05 \mathrm{~g/mol} = 9.615 \mathrm{~g}.\]

Step 6: Match Calculation with Answer Choices

From the calculation, the closest answer by rounding is 9.6 g. Thus, the correct option is (a) \(9.6 \mathrm{~g}\).

Key Concepts

Solution Preparation

Creating a chemical solution is like cooking, you need the correct amounts of each ingredient. Here, the solution preparation involves mixing a certain amount of a chemical solute, in this case, methanol (\(\mathrm{CH}_{3} \mathrm{OH}\)), with a solvent, typically water, to achieve a desired concentration, referred to as molarity.

To prepare a specific molarity:

• Understand that molarity (M) is defined as moles of solute per liter of solution.
• Concentration can be visualized as how packed a solution is with solute particles.
• In this exercise, you aim to make a 2.0 M solution, meaning there should be 2 moles of methanol in one liter of the solution.

The volume of your desired solution here is given as 150 mL. It's crucial to convert this to liters by using the conversion factor that 1 liter equals 1000 mL, leading to 0.150 L. Remember, molarity calculations always deal with liters for volume.

Moles Calculation

Calculating moles is essential because solutions are most easily measured by their moles instead of grams. This step involves determining how many moles of your solute, methanol in this exercise, you will need for your solution.

Here, with a goal of preparing a 2.0 M methanol solution in 0.150 L, use the formula:

• Number of moles = Molarity \( \times \) Volume in Liters
• Insert the desired molarity: 2.0 M
• And the volume in liters: 0.150 L

So, you will perform the multiplication: 2.0 M \( \times \) 0.150 L, which results in 0.300 moles of methanol. This means you will need precisely 0.300 moles of \(\mathrm{CH}_{3} \mathrm{OH}\) to prepare your solution.

Molar Mass Calculation

The last key step involves converting the calculated moles of methanol to grams. This requires the molar mass, which tells you how much one mole of methanol weighs.

To find the molar mass of \(\mathrm{CH}_{3} \mathrm{OH}\):

• Add the atomic masses of all the atoms in one molecule: one carbon (12.01 g/mol), four hydrogens (4 \(\times\) 1.01 g/mol), and one oxygen (16.00 g/mol).
• The sum is 32.05 g/mol.

With the molar mass found, utilize it in the conversion:

• Grams needed = Moles calculated \( \times \) Molar mass
• This means: 0.300 moles \( \times \) 32.05 g/mol = 9.615 grams of \(\mathrm{CH}_{3} \mathrm{OH}\)

Finally, choose the answer closest to your calculation from the options provided. In this case, it's 9.6 g, making option (a) correct.