Q.The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. $K_b$ for benzene is 2.53 K kg mol$^{-1}$.

This problem involves the colligative property of Boiling Point Elevation. The addition of a non-volatile solute raises the boiling point of a solvent.

First, calculate the change in boiling point ($\Delta T_b$):

Next, use the boiling point elevation formula to find the molality ($m$) of the solution:

Rearranging for $m$:

Now, calculate the moles of solute ($n_2$) using the definition of molality and the mass of solvent (90 g = 0.090 kg):

$n_2 = m \times \text{mass of solvent (kg)} = 0.347826 \text{ mol kg}^{-1} \times 0.090 \text{ kg} = 0.031304 \text{ mol}$ …

When a non-volatile solute is dissolved in a pure solvent, the vapor pressure of the solvent decreases. This is because the solute particles occupy some of the surface area, reducing the number of solvent molecules that can escape into the vapor phase. For the solution to boil, its vapor pressure must reach the external atmospheric pressure. Since the vapor pressure is now lower at any given temperature, a higher temperature is required to achieve the boiling point. This phenomenon is known as boiling point elevation, and it is a colligative property, meaning it depends only on the number of solute particles, not their identity.

The extent of boiling point elevation ($\Delta T_b$) is directly proportional to the molality ($m$) of the solution:

We can use this relationship to find the molar mass of the unknown solute.

1. Identify the given information and the goal.

   We are given:

   • Boiling point of pure benzene ($T_b^0$) = $353.23 \text{ K}$
   • Boiling point of the solution ($T_b$) = $354.11 \text{ K}$
   • Mass of solute ($w_2$) = $1.80 \text{ g}$
   • Mass of solvent (benzene, $w_1$) = $90 \text{ g}$
   • Molal elevation constant for benzene ($K_b$) = $2.53 \text{ K kg mol}^{-1}$ Our goal is to calculate the molar mass of the solute ($M_2$).

2. Calculate the boiling point elevation ($\Delta T_b$).

   The elevation in boiling point is the difference between the boiling point of the solution and the boiling point of the pure solvent.

3. Calculate the molality ($m$) of the solution. Using the boiling point elevation formula:

We can rearrange this to solve for molality:

Substitute the values:

4. Convert the mass of the solvent to kilograms. Molality is defined as moles of solute per kilogram of solvent. The given mass of benzene is in grams, so we convert it to kilograms: …

Method: Boiling Point Elevation (Using the Formula $\Delta T_b = K_b \cdot m$)

This method uses the direct relationship between the elevation in boiling point and the molality of the solution.

Steps

1. Calculate the elevation in boiling point ($\Delta T_b$)

The boiling point of pure benzene is $353.23\ \text{K}$, and the solution boils at $354.11\ \text{K}$.

So,

2. Write the boiling point elevation formula

where $K_b = 2.53\ \text{K kg mol}^{-1}$ and $m$ is the molality of the solution.

3. Express molality in terms of solute mass and molar mass

Molality $m$ is moles of solute per kg of solvent.

Let $M$ be the molar mass of the solute (in g/mol).

Moles of solute = $\dfrac{1.80}{M}$

Mass of benzene = $90\ \text{g} = 0.090\ \text{kg}$

Thus, …

This is a classic Boiling Point Elevation problem from physical chemistry, often seen in CBSE, JEE, and other Indian exams.

✓ The Correct Approach (Quick Recap)

We use:

Where:

• $\Delta T_b = T_b(\text{solution}) - T_b(\text{solvent}) = 354.11 - 353.23 = 0.88\ \text{K}$
• $K_b = 2.53\ \text{K kg mol}^{-1}$
• $m = \text{molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$

Let $M$ = molar mass of solute (g/mol).

Then:

Final answer: $\mathbf{58\ g\ mol^{-1}}$ — the fraction evaluates to 57.5, which NCERT rounds to two significant figures and prints as 58.

✗ Common Mistakes & How to Avoid Them

1. Forgetting to convert solvent mass to kg

• Mistake: Using 90 g directly in the formula without converting to 0.090 kg.
• Why it's wrong: Molality is moles per kg of solvent — using grams gives an answer off by a factor of 1000.
• How to avoid: Always write the unit: mass of solvent in kg. If given in grams, divide by 1000 immediately.

2. Using mass of solution instead of mass of solvent

• Mistake: Taking 90 g + 1.80 g = 91.8 g as the denominator.
• Why it's wrong: Molality uses solvent mass only, not solution mass.
• How to avoid: Read carefully — the problem says "in 90 g of benzene". That's the solvent mass.

3. Confusing $\Delta T_b$ with the final boiling point

• Mistake: Plugging 354.11 K directly into the formula as $\Delta T_b$.
• Why it's wrong: $\Delta T_b$ is the change (final − initial), not the final value.
• How to avoid: Always compute $\Delta T_b = T_{\text{solution}} - T_{\text{solvent}}$ first, then use it.

4. Unit mismatch in $K_b$

• Mistake: Using $K_b$ without checking that it matches the units of $\Delta T_b$ (K) and mass (kg).
• Why it's wrong: $K_b$ is in K kg mol⁻¹ — if you use grams, the units won't cancel.
• How to avoid: Keep all quantities in SI units (kg for mass, K for temperature).

5. Rounding too early in the working

• Mistake: Rounding $\Delta T_b$ to 0.9 K, or rounding molality to one digit, before the final division.
• Why it's wrong: Rounding an intermediate quantity compounds the error — using $\Delta T_b = 0.9$ K, for example, gives $M \approx 56$ g/mol instead of the correct 57.5. …