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 approximately 253 g/mol.

Step 1: Identify Known Values

We have a mass of solute, \(m = 10 \text{ g}\), and a mass of solvent (benzene), \( W = 100 \text{ g} = 0.1 \text{ kg} \). The boiling point elevation, \( \Delta T_b = 1^{\circ} \text{C} \), and the ebullioscopic constant for benzene, \( K_b = 2.53 \, \text{K} \, \text{m}^{-1} \). We need to find the molecular mass, \( M \), of the solute.

Step 2: Apply the Boiling Point Elevation Formula

The formula for boiling point elevation is \( \Delta T_b = K_b \times m \), where \( m \) is the molality of the solution. Molality \( m = \frac{n}{W} \), with \( n \) being the number of moles of solute.

Step 3: Relate Molality to Molecular Mass

From the boiling point elevation formula, rearrange to find: \[ m = \frac{\Delta T_b}{K_b} \]. Since \( m = \frac{n}{W} \) and \( n = \frac{m}{M} \), substitute: \( \frac{m}{M} = \frac{\Delta T_b}{K_b} \times W \).

Step 4: Solve for Molecular Mass

Plug in values into the equation: \[ \frac{10}{M} = \frac{1}{2.53} \times 0.1 \]. Calculate: \[ \frac{10}{M} = \frac{0.1}{2.53} = 0.03952 \]. So, \( M = \frac{10}{0.03952} = 253.02 \text{ g/mol} \).

Step 5: Choose the Nearest Answer

The calculated molecular mass of the solute is approximately \( 253 \text{ g/mol} \). Compare this with the provided options, and select the closest one, which is (d) \( 253 \text{ g} \).

Key Concepts

Non-volatile solute

When a non-volatile solute is added to a solvent, like benzene, it affects the boiling point of the solvent. A non-volatile solute doesn't vaporize easily. Instead, it stays dissolved in the solution. Because of this, it modifies the physical properties of the solvent. This is a phenomenon known as "boiling point elevation."

The non-volatile solute causes an increase in the boiling point of the solvent. It requires more energy (higher temperature) to break the interactions between the molecules in the solution. This is why, when you dissolve something like a non-volatile solute in benzene, the observed boiling point of benzene increases.

• This increase depends on the amount of solute dissolved.
• It doesn't depend on the type of solute, hence it's a colligative property.
• This is crucial for applications like antifreeze mixtures where the boiling or freezing points have to be adjusted.

Molecular mass calculation

Molecular mass calculation is a vital method to determine how much one mole of a given substance weighs in grams. For calculating the molecular mass of a solute in a solution, one can leverage the concept of boiling point elevation.

Start with the formula: \[\Delta T_b = K_b \times m\]Here, \( \Delta T_b \) is the boiling point elevation, and \( K_b \) is the ebullioscopic constant specific to the solvent. Molality \( m \), a measure of concentration, is calculated as:\[m = \frac{n}{W}\]
The number of moles \( n \) of the solute can be given by the formula:\[n = \frac{m}{M} \]where \( m \) is the mass of the solute and \( M \) is the molecular mass you need to find.

To find \( M \), rearrange the formula:\[M = \frac{m \times K_b \times W}{\Delta T_b}\]By using the provided information and solving this equation, you can determine the molecular mass of the solute, which tells you how heavy a single molecule or formula unit is.

Ebullioscopic constant for benzene

The ebullioscopic constant \( K_b \) is a proportionality factor that is unique to each solvent. It indicates how much the boiling point of the solvent will increase as a result of the addition of one mole of a non-volatile solute to 1 kg of the solvent.

• This property varies between substances, making it an important figure for calculating the effects of solutes.
• The unit is Kelvin per molality (K m^-1).

For benzene, \( K_b \) is 2.53 K m^-1. This constant helps in calculating how the boiling point will elevate based on the quantity of the solute introduced.

For instance, if you know how much the boiling point has increased due to a specific amount of solute, you can use the \( K_b \) to work backwards, identify the molality of the solution, and eventually determine the molecular mass of the solute. This constant is integral in performing such calculations accurately and understanding the extent of changes in the solvent's boiling point due to the solute presence.