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 correct answer is (a) 9.6 g.

Step 1: Understand Molarity

Molarity (M) is defined as the number of moles of solute divided by the liters of solution. In this case, we need to calculate how many moles of methanol (\(\mathrm{CH}_{3} \mathrm{OH}\)) are needed for a \(2.0 \mathrm{M}\) solution in \(150 \mathrm{~mL}\) of solution.

Step 2: Convert Volume to Liters

Since molarity is moles per liter, convert the volume from milliliters to liters. \[150 \mathrm{~mL} = 0.150 \mathrm{~L}\]

Step 3: Calculate Moles of Solute

Multiply the molarity by the volume in liters to find the moles of \(\mathrm{CH}_{3} \mathrm{OH}\). \[\text{Moles of } \mathrm{CH}_{3} \mathrm{OH} = 2.0 \mathrm{M} \times 0.150 \mathrm{~L} = 0.300 \text{ moles}\]

Step 4: Calculate Molar Mass of Methanol

The molar mass of \(\mathrm{CH}_{3} \mathrm{OH}\) is the sum of the atomic masses of all the atoms in a molecule: \text{C} (12 \mathrm{~g/mol}), \text{H} (1 \mathrm{~g/mol} \times 4) and \text{O} (16 \mathrm{~g/mol}). \[\text{Molar mass} = 12 + (4 \times 1) + 16 = 32 \mathrm{~g/mol}\]

Step 5: Convert Moles to Grams

Multiply the number of moles by the molar mass to find the mass in grams.\[\text{Mass of } \mathrm{CH}_{3} \mathrm{OH} = 0.300 \text{ moles} \times 32 \mathrm{~g/mol} = 9.6 \mathrm{~g}\]

Step 6: Choose the Correct Option

The calculated mass of \(\mathrm{CH}_{3} \mathrm{OH}\) is \(9.6 \mathrm{~g}\). So, the correct option is (a) \(9.6 \mathrm{~g}\).

Key Concepts

Solution Preparation

When preparing a chemical solution, it's critical to know the desired concentration and volume you need. This exercise focuses on creating a solution with a specific molarity and volume of \(150 \, \text{mL}\).
The solution preparation starts with accurately measuring the desired volume of solvent. In this example, we're working with water as the solvent. Using precise instruments like graduated cylinders can help ensure you achieve the exact volume. Next, weigh the required amount of solute, which in our case is methanol (\(\text{CH}_{3}\text{OH}\)) using a precise scale. This involves understanding and calculating the required moles based on the molarity formula.
The formula \[\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}}\] guides the calculation of the right amount of substance to dissolve, ensuring that the solution has the correct concentration.
After determining the moles, converting them into grams allows you to physically measure out the solute with the appropriate precision. Dissolve the solute in the solvent, often requiring some stirring to ensure complete dissolution, resulting in the final prepared solution.

Mole Calculations

Understanding how to perform mole calculations is crucial in chemistry, especially when preparing solutions. Moles give chemists a way to count particles using Avogadro’s number, but in practice, moles are often calculated from the mass of a substance based on its molar mass.
In a typical exercise like converting a desired molarity to grams of a substance, start by ensuring your volume is in liters. Molarity (\(\text{M}\)) is moles per liter, so converting your volume is key. This exercise moves from a given volume in milliliters (\(150 \, \text{mL}\)) to liters (\(0.150 \, \text{L}\)) to correctly use the molarity.

• Step 1: Calculate the moles of solute required using the molarity equation. Here, this requires multiplying the molarity by volume (\(2.0 \, \text{M} \times 0.150 \, \text{L} = 0.300 \, \text{moles}\)).
• Step 2: Know the molar mass of methanol, which involves summing the atomic masses of its constituent atoms: \(\text{C (12 g/mol)}, \text{H (1 g/mol)} \times 4, \text{O (16 g/mol)}\). Therefore, the molar mass is \(32 \, \text{g/mol}\).
• Step 3: Translate those moles into grams using the molar mass: \(0.300 \, \text{moles} \times 32 \, \text{g/mol} = 9.6 \, \text{g}\).

These steps are essential to converting theoretical chemical equations into practical laboratory work.

Conversion of Units

Unit conversion in chemistry is the bridge that connects theoretical calculations to practical execution. Being able to switch between units correctly ensures that solutions and compounds are prepared accurately.
One such conversion illustrated in this exercise involves converting mL to L, a fundamental step for using molarity. Given \(150 \, \text{mL}\), we convert it to liters by recognizing that 1000 mL equals 1 L. Thus, \[150 \, \text{mL} = 0.150 \, \text{L}\]
Conversions also necessitate understanding molar mass units in grams per mole (\(\text{g/mol}\)), emphasized when translating moles of a substance into grams for practical measurements. This requires multiplying the number of moles by the molar mass to find the mass in grams. Being precise in these conversions helps prevent errors in solution preparation, ensuring reliable results and analyses.
Remember, the key to mastering chemistry is maintaining consistency in units throughout all calculations.