Question

The vapour pressure of pure benzene at a certain temperature is 0.85 bar. When 0.5 g of a non-volatile solute is added to 39 g of benzene, the vapour pressure becomes 0.845 bar. What is the molar mass of the substance? ________.

• A. $85g~mol^{-1}$
• B. $127.5~g~mol^{-1}$
• C. $170~g~mol^{-1}$
• D. $210g~mol^{-1}$
• E. $145~g~mol^{-1}$

Hint:

For dilute solutions, $\frac{\Delta P}{P^{\circ}} \approx \frac{w_2 M_1}{M_2 w_1}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
This problem deals with the colligative property of vapor pressure lowering. According to Raoult's law, when a non-volatile solute is dissolved in a solvent, the vapor pressure of the solvent is lowered. The extent of this lowering depends on the mole fraction of the solute. We can use this relationship to determine the molar mass of the solute.
Step 2: Key Formula or Approach
Raoult's Law states that the relative lowering of vapor pressure is equal to the mole fraction of the solute (\(\chi_{solute}\)). \[ \frac{P^{\circ}_{solvent} - P_{solution}}{P^{\circ}_{solvent}} = \chi_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}} \] For dilute solutions, this can be approximated as: \[ \frac{P^{\circ}_{solvent} - P_{solution}}{P^{\circ}_{solvent}} \approx \frac{n_{solute}}{n_{solvent}} \] where \(n = \frac{\text{mass}}{\text{molar mass}}\). We will use this approximation to solve for the molar mass of the solute.
Step 3: Detailed Explanation
1. Identify and list the given information.
- Pure solvent (benzene) vapor pressure, \(P^{\circ} = 0.85 \text{ bar}\) - Solution vapor pressure, \(P = 0.845 \text{ bar}\) - Mass of solvent (benzene), \(w_{solvent} = 39 \text{ g}\) - Molar mass of solvent (benzene), \(M_{solvent} = 78 \text{ g/mol}\) - Mass of solute, \(w_{solute} = 0.5 \text{ g}\) - Molar mass of solute, \(M_{solute}\) = ? 2. Calculate the moles of solvent (benzene).
\[ n_{solvent} = \frac{w_{solvent}}{M_{solvent}} = \frac{39 \text{ g}}{78 \text{ g/mol}} = 0.5 \text{ mol} \] 3. Apply the dilute solution form of Raoult's Law.
\[ \frac{P^{\circ} - P}{P^{\circ}} \approx \frac{n_{solute}}{n_{solvent}} = \frac{w_{solute}/M_{solute}}{n_{solvent}} \] Substitute the values: \[ \frac{0.85 - 0.845}{0.85} = \frac{0.5 / M_{solute}}{0.5} \] \[ \frac{0.005}{0.85} = \frac{1}{M_{solute}} \] \[ \frac{5}{850} = \frac{1}{M_{solute}} \] \[ \frac{1}{170} = \frac{1}{M_{solute}} \] 4. Solve for \(M_{solute}\).
\[ M_{solute} = 170 \text{ g/mol} \] Step 4: Final Answer
The molar mass of the substance is 170 g mol\(^{-1}\).