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

Add 9.6 g of \(\mathrm{CH}_3\mathrm{OH}\) to prepare a 2.0 M solution.

Step 1: Understanding Molarity

Molarity (M) is defined as the number of moles of solute per liter of solution. Here, the solution is 2.0 M, meaning there are 2.0 moles of \( \mathrm{CH}_{3}\mathrm{OH} \) per liter of solution.

Step 2: Convert Volume to Liters

First, convert the solution volume from milliliters to liters because molarity involves liters. Since there are 1000 mL in a liter, \( 150 \text{ mL} = 0.150 \text{ L} \).

Step 3: Calculate Moles Needed

The formula for moles using molarity is: \( \text{Moles} = \text{Molarity} \times \text{Volume (L)} \) Use \( 2.0 \text{ M} \) and \( 0.150 \text{ L} \). \[ \text{Moles of } \mathrm{CH}_{3}\mathrm{OH} = 2.0 \times 0.150 = 0.300 \text{ mol} \]

Step 4: Find Molar Mass of \(\mathrm{CH}_3\mathrm{OH}\)

Calculate the molar mass of \( \mathrm{CH}_3\mathrm{OH} \) by summing the atomic masses: - C: 12.01 g/mol - H: 1.01 g/mol (there are 4 H atoms) - O: 16.00 g/mol \[ \text{Molar Mass of } \mathrm{CH}_3\mathrm{OH} = 12.01 + 4(1.01) + 16.00 = 32.05 \ \text{g/mol} \]

Step 5: Calculate Grams of \(\mathrm{CH}_3\mathrm{OH}\) Needed

Use the relationship: \( \text{Mass} = \text{Moles} \times \text{Molar Mass} \). Thus, the mass of \( \mathrm{CH}_3\mathrm{OH} \) needed is: \[ \text{Mass} = 0.300 \times 32.05 = 9.615 \text{ grams} \]This rounds to \(9.6 \text{ grams} \).

Step 6: Choose the Correct Answer from Options

The calculated mass of \( \mathrm{CH}_3\mathrm{OH} \) needed is \(9.6 \text{ grams} \), which corresponds to option (a).

Key Concepts

Understanding Molarity

Molarity is a concept that relates to the concentration of a solute in a solution. It is denoted by the symbol \(M\), and it's defined as the number of moles of a solute per liter of solution. Think of it as a way to measure how crowded the solute particles are in a given volume of liquid. When we say a solution is 2.0 M \(\mathrm{CH}_3\mathrm{OH}\), it means there are 2.0 moles of \(\mathrm{CH}_3\mathrm{OH}\) in every liter of the solution.

• Understanding molarity helps us predict how solutions will behave in chemical reactions, since reactions depend on the number of moles of reactants.
• Thus, knowing the molarity allows chemists to use volumes of solutions to calculate the amounts of substances involved in reactions.

Molarity is foundational for calculations in chemistry because it connects the physical measurement of volume with the chemical counting measure of moles. It’s essential in fields like analytical chemistry, where precise concentrations are crucial for reactions.

Calculating Molar Mass

Molar mass is crucial when converting between the mass of a substance and the moles of a substance. It is the mass in grams of one mole of a chemical compound and is expressed in units of grams per mole (g/mol). Calculating the molar mass involves summing the atomic masses of all atoms present in a molecule.
For example, to calculate the molar mass of \(\mathrm{CH}_3\mathrm{OH}\):

• Carbon (C) has an atomic mass of 12.01 g/mol.
• Hydrogen (H), with four atoms in \(\mathrm{CH}_3\mathrm{OH}\), each has an atomic mass of 1.01 g/mol, totaling 4.04 g/mol for hydrogen.
• Oxygen (O) contributes 16.00 g/mol.

Adding these together gives a molar mass of 32.05 g/mol for \(\mathrm{CH}_3\mathrm{OH}\). Knowing the molar mass allows you to convert between mass and moles easily using the formula: \[\text{Mass} = \text{Moles} \times \text{Molar Mass}\]This relationship is extremely useful for preparing specific amounts of solutions, as we've seen when determining how many grams of \(\mathrm{CH}_3\mathrm{OH}\) are needed.

Volume Conversion in Chemistry

In chemistry, accurately converting and working with volumes is vital, especially when dealing with molarity calculations. Since molarity is defined in terms of liters, it is often necessary to convert smaller volumes to liters because practical laboratory measurements may start in milliliters.
Volume conversion requires knowing the basic relationship that there are 1000 milliliters in a liter. This helps us convert volumes easily. For example:

• To convert 150 mL to liters, use the following calculation: \(150 \text{ mL} \div 1000 = 0.150 \text{ L} \).
• This step ensures that when we multiply by molarity, we're using consistent units, important for accurate calculations.

Regular practice of volume conversions helps you perform chemical calculations smoothly without errors. Converting volumes ensures you apply the correct amount of solute needed based on molarity, keeping your solutions accurate and effective.