Problem 55 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

A sugar syrup of weight \(214.2 \mathrm{~g}\) contains \(34.2 \mathrm{~g}\) of sugar \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\). Calculate : (i) molal concentration and (ii) mole fraction of sugar in the syrup.

Step-by-Step Solution

Answer

(i) Molality = 0.56 mol/kg, (ii) Mole fraction of sugar = 0.01

Step 1: Understanding Molality

Molality is defined as the number of moles of solute (sugar in this case) per kilogram of solvent. The formula for molality \(m\) can be expressed as \( m = \frac{n}{m_{\text{solvent}}} \), where \(n\) is the moles of solute and \(m_{\text{solvent}}\) is the mass of the solvent in kilograms.

Step 2: Calculate Moles of Sugar

First, calculate the moles of sugar \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\). The molar mass of sugar is approximately \(342.3 \, \mathrm{g/mol}\). Thus, the number of moles of sugar is given by \( n = \frac{34.2 \, \mathrm{g}}{342.3 \, \mathrm{g/mol}} \approx 0.10 \, \mathrm{mol} \).

Step 3: Calculate Mass of Solvent

Subtract the mass of sugar from the total mass of the syrup to get the mass of the solvent (water). \( 214.2 \, \mathrm{g} - 34.2 \, \mathrm{g} = 180.0 \, \mathrm{g} \). Convert this mass into kilograms: \( 180.0 \, \mathrm{g} = 0.180 \, \mathrm{kg} \).

Step 4: Calculate Molality

Use the molality formula: \( m = \frac{0.10 \, \mathrm{mol}}{0.180 \, \mathrm{kg}} \approx 0.56 \, \mathrm{mol/kg} \). Thus, the molal concentration of the sugar in the syrup is \(0.56 \, \mathrm{mol/kg} \).

Step 5: Understanding Mole Fraction

The mole fraction is defined as the ratio of the moles of a component to the total moles of all components in the mixture. For sugar, \( X_{\text{sugar}} = \frac{n_{\text{sugar}}}{n_{\text{sugar}} + n_{\text{water}}} \).

Step 6: Calculate Moles of Water

Calculate the moles of water. The molar mass of water is \(18.02 \, \mathrm{g/mol}\). Thus, \( n_{\text{water}} = \frac{180.0 \, \mathrm{g}}{18.02 \, \mathrm{g/mol}} \approx 9.99 \, \mathrm{mol} \).

Step 7: Calculate Mole Fraction of Sugar

Using the mole fraction formula: \( X_{\text{sugar}} = \frac{0.10}{0.10 + 9.99} \approx 0.01 \). Thus, the mole fraction of sugar in the syrup is approximately \(0.01\).

Key Concepts

Mole Fraction

The mole fraction is a concept used to describe the proportion of a component within a mixture. For example, in a syrup mixture, we want to understand how much of the mixture is made up of sugar. This is expressed through the mole fraction.

To calculate the mole fraction, we need the moles of sugar and the total moles of the solution (sugar and water, in this case). You calculate the mole fraction of a component by dividing the number of moles of that component by the total number of moles of all components.

The formula is straightforward:

• Mole fraction of sugar: \( X_{\text{sugar}} = \frac{n_{\text{sugar}}}{n_{\text{sugar}} + n_{\text{water}}} \)

In the syrup example, we calculate:

• Moles of sugar: 0.10 mol
• Moles of water: 9.99 mol
• Total moles: 0.10 + 9.99 = 10.09 mol
• Thus, \( X_{\text{sugar}} \approx 0.01 \)

By understanding the mole fraction, you can easily find out the proportion of each component in a mixture.

Moles Calculation

Calculating moles is essential for determining various concentration metrics, such as mole fraction and molality. The mole is a fundamental unit in chemistry which represents a quantity of particles – typically atoms or molecules.

To find the number of moles, use the formula:

• \( n = \frac{m}{M} \)

where:

• \( n \) is the number of moles,
• \( m \) is the mass of the substance in grams, and
• \( M \) is the molar mass of the substance in grams per mole (g/mol).

For sugar, with a molar mass of approximately 342.3 g/mol, and a given mass of 34.2 g, the calculation becomes:

• \( n = \frac{34.2 \, \text{g}}{342.3 \, \text{g/mol}} \approx 0.10 \, \text{mol} \)

Similarly, for water with a molar mass of 18.02 g/mol, and a mass of 180.0 g:

• \( n = \frac{180.0 \, \text{g}}{18.02 \, \text{g/mol}} \approx 9.99 \, \text{mol} \)

Using these calculations helps to understand and quantify the components of a mixture.

Concentration Calculations

Concentration calculations help us express the amount of solute in a solution, which can be crucial for understanding a mixture's properties. This involves determining how much of a given substance is present compared to the entire mixture.

In this exercise, we've considered molal concentration which is particularly useful when dealing with temperature-dependent properties, such as boiling point elevation or freezing point depression. **Molality** is measured in moles of solute per kilogram of solvent, distinct from molarity which measures moles per liter of solution.

The formula is:

• \( m = \frac{n}{m_{\text{solvent}}} \)

where:

• \( n \) is the moles of solute,
• \( m_{\text{solvent}} \) is the mass of the solvent in kilograms.

For sugar syrup, with calculated moles of sugar as 0.10 mol and water weighing 0.180 kg, use:

• \( m = \frac{0.10}{0.180} \approx 0.56 \, \text{mol/kg} \)

This tells us the molal concentration is approximately 0.56 molal, indicating the density of sugar in the syrup solution.