Problem 81 - A Complete Resource Book in Chemistry for JEE Main

Question

\(5 \mathrm{~g}\) of \(\mathrm{CH}_{3} \mathrm{COOH}\) is dissolved in one litre of ethanol. Suppose there is no reaction between them. If the density of ethanol is \(0.789 \mathrm{~g} / \mathrm{mL}\) then the molality of resulting solution is (a) \(0.0256\) (b) \(0.1056\) (c) \(1.1288\) (d) \(0.2076\)

Step-by-Step Solution

Answer

The molality of the solution is \( 0.1056 \), option (b).

Step 1: Understanding Molality

The molality of a solution is defined as the number of moles of solute per kilogram of solvent. It is represented as \[ m = \frac{n}{m_{\text{solvent}}} \] where \( n \) is the number of moles of the solute and \( m_{\text{solvent}} \) is the mass of the solvent in kilograms.

Step 2: Calculate the Moles of Solute

Firstly, we need to calculate the number of moles of acetic acid \( \mathrm{CH}_3 \mathrm{COOH} \). The molar mass of \( \mathrm{CH}_3 \mathrm{COOH} \) is approximately \( 60.05 \text{ g/mol} \). So, the number of moles \( n \) is given by:\[ n = \frac{5 \text{ g}}{60.05 \text{ g/mol}} \approx 0.0833 \text{ mol} \]

Step 3: Determine the Mass of Ethanol Solvent

Given the density of ethanol is \( 0.789 \text{ g/mL} \) and there are \( 1000 \text{ mL} \) in \( 1 \text{ L} \), the mass of ethanol is:\[ \text{Mass of ethanol} = 1000 \text{ mL} \times 0.789 \text{ g/mL} = 789 \text{ g} \] Convert this mass to kilograms for molality calculation:\[ m_{\text{solvent}} = \frac{789 \text{ g}}{1000} = 0.789 \text{ kg} \]

Step 4: Calculate the Molality of the Solution

Now, substitute the values of moles \( n \) and solvent mass \( m_{\text{solvent}} \) into the formula for molality:\[ m = \frac{0.0833 \text{ mol}}{0.789 \text{ kg}} \approx 0.1056 \text{ mol/kg} \]

Step 5: Choose the Correct Option

Match the calculated molality to the given options. The computed molality \( 0.1056 \text{ mol/kg} \) corresponds to option (b) \( 0.1056 \).

Key Concepts

Moles of Solute

To begin understanding molality, the first step is to calculate the moles of the solute. A solute is the substance that is dissolved in a solution. The chemical acetic acid, symbolized as \( \mathrm{CH}_3 \mathrm{COOH} \), is the solute in this exercise. We must determine how many moles of \( \mathrm{CH}_3 \mathrm{COOH} \) are present in the given 5 grams.
The calculation starts by knowing the molar mass, which is the weight of one mole of a substance. For acetic acid, the molar mass is approximately \( 60.05 \text{ g/mol} \).

• Formula: \( n = \frac{\text{mass of solute}}{\text{molar mass}} \)
• Substituting the values: \( n = \frac{5 \text{ g}}{60.05 \text{ g/mol}} \)
• Calculated moles: approximately \( 0.0833 \text{ mol} \)

This tells us the number of moles of acetic acid present, a crucial step in determining molality.

Density of Ethanol

Density is an essential factor in the calculation since it helps us figure out the mass of the solvent from its volume. Ethanol is the solvent in this solution, meaning it is the component in greater amount, that dissolves the solute. The density of ethanol in this problem is provided as \(0.789 \mathrm{~g} / \mathrm{mL}\). Knowing that there are \(1000 \text{ mL}\) in \(1 \text{ L}\), we can calculate the mass of ethanol by multiplying its density by the number of milliliters.

• Formula: \( \text{Mass of ethanol} = \text{Volume of ethanol} \times \text{Density of ethanol} \)
• Substitute values: \( \text{Mass of ethanol} = 1000 \text{ mL} \times 0.789 \text{ g/mL} \)
• Calculated mass: \(789 \text{ g}\)

Converting this mass into kilograms is essential for later use in the molality formula.

Mass of Solvent

With the mass of ethanol calculated as 789 grams, we need to convert it into kilograms, as molality's definition requires the mass of the solvent to be in kilograms. Conversion to kilograms involves a simple step of dividing grams by 1000.

• Conversion: \( m_{\text{solvent}} = \frac{789 \text{ g}}{1000} \)
• Result: \(0.789 \text{ kg}\)

This value represents the correct mass of ethanol solvent in the molality equation, setting the stage for calculating the final concentration of the solution.

Calculation of Molality

Finally, with the moles of the solute and the mass of the solvent known, we can use the formula for molality. Molality is and often used when temperature conditions fluctuate in experiments since it does not change with temperature. The formula for molality is \( m = \frac{n}{m_{\text{solvent}}} \). With the moles of solute as \(0.0833 \text{ mol}\) and the mass of solvent as \(0.789 \text{ kg}\), we substitute these values in:

• \( m = \frac{0.0833 \text{ mol}}{0.789 \text{ kg}} \)
• Calculated molality: approximately \(0.1056 \text{ mol/kg}\)

This molality value of \(0.1056 \text{ mol/kg}\) directly corresponds to option (b) in the original question, providing the final answer to the exercise.