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

Question

By dissolving \(10 \mathrm{~g}\) of a non-volatile solute in \(100 \mathrm{~g}\) of benzene, the boiling point rises by \(1^{\circ} \mathrm{C}\). The molecular mass of solute is \(\left[\mathrm{K}_{\mathrm{b}}\right.\) for benzene \(\left.=2.53 \mathrm{~K} \mathrm{~m}^{-1}\right]\) (a) \(235 \mathrm{~g}\) (b) \(352 \mathrm{~g}\) (c) \(250 \mathrm{~g}\) (d) \(253 \mathrm{~g}\)

Step-by-Step Solution

Answer

The molecular mass of the solute is 253 g/mol, corresponding to option (d).

Step 1: Understanding the boiling point elevation formula

The formula for boiling point elevation is given by \[\Delta T_b = i \cdot K_b \cdot m\]where \( \Delta T_b \) is the change in boiling point, \( K_b \) is the ebullioscopic constant, \( m \) is the molality, and \( i \) is the van't Hoff factor (which is 1 for non-volatile solutes). In this problem, \( \Delta T_b = 1^{\circ} \mathrm{C} \) and \( K_b = 2.53 \mathrm{~K} \mathrm{~m}^{-1} \).

Step 2: Calculating the molality

Molality \( m \) is calculated as moles of solute per kilogram of solvent. First, rewrite the boiling point formula as: \[m = \frac{\Delta T_b}{K_b}\]Substitute the known values: \[m = \frac{1}{2.53} = 0.395\, \mathrm{mol/kg}\]

Step 3: Calculating the moles of solute

Moles of solute can be calculated using molality \[m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}\]Rearranging gives: \[\text{moles of solute} = m \times \text{mass of solvent in kg}\]Substituting the values, \[\text{mass of solvent in kg} = \frac{100\,\text{g}}{1000} = 0.1\, \text{kg}\]\[\text{moles of solute} = 0.395 \times 0.1 = 0.0395\, \text{mol}\]

Step 4: Finding the molecular mass of the solute

Molecular mass is calculated as: \[\text{Molecular mass} = \frac{\text{mass of solute in g}}{\text{moles of solute}}\]Substitute the values: \[\text{Molecular mass} = \frac{10}{0.0395} \approx 253 \mathrm{~g/mol}\]

Step 5: Determining the correct option

The calculated molecular mass of the solute is approximately 253 g/mol. Comparing it with the given options, the correct answer is (d) 253 g/mol.

Key Concepts

Non-volatile Solute

In the context of this exercise, a non-volatile solute refers to a substance that does not readily evaporate into a gas under existing conditions. When a non-volatile solute is dissolved in a solvent like benzene, it increases the boiling point of the solvent. This phenomenon is known as boiling point elevation. Why does this happen? Because the presence of the solute decreases the solvent's vapor pressure, requiring a higher temperature to achieve boiling.

It's essential to note that for non-volatile solutes, the van't Hoff factor, which represents how many particles the solute breaks into in solution, is commonly 1. This simplifies calculations because the solute doesn't dissociate into smaller particles. In this problem, the boiling point of benzene rises by 1°C due to the dissolved non-volatile solute, demonstrating the impact of such solutes on boiling point elevation.

Molality

Molality is a key concentration unit in chemistry, particularly when discussing colligative properties like boiling point elevation. It measures the number of moles of solute per kilogram of solvent. Unlike molarity, which depends on volume, molality heavily relies on mass, making it unaffected by temperature changes. This characteristic makes it particularly useful in boiling point and freezing point scenarios.

In this exercise, we used the boiling point elevation formula to calculate molality. Knowing the change in boiling point (
°C) and the ebullioscopic constant (.53 K m) for benzene, we could determine the molality by dividing the temperature change by the constant. This calculation resulted in a molality of 0.395 mol/kg. This value signifies how concentrated the solute is within the benzene and directly contributes to the observed increase in boiling point.

Molecular Mass Determination

Determining the molecular mass of a solute is crucial in identifying the substance and understanding its properties. The molecular mass is essentially the mass of one mole of a given compound. In this exercise, after finding the molality and moles of solute, we could calculate the molecular mass using the formula:

1. Molecular mass = mass of solute / moles of solute

By knowing that we have 10 grams of the solute and 0.0395 moles, we find the molecular mass to be approximately 253 g/mol. This exact value helps determine the specific non-volatile solute dissolved in benzene. Understanding how to perform these calculations is vital for any chemistry student, as it bridges theoretical concepts with practical applications in laboratory settings.