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Step-by-Step Solution
Step 1: Identify the Given Data
• Vapour pressure of pure acetone at 20°C, $P^\circ = 185\ \text{torr}$
• Vapour pressure of the solution at 20°C, $P_s = 183\ \text{torr}$
• Mass of solute (non-volatile substance), $w_2 = 1.2\ \text{g}$
• Mass of solvent (acetone), $w_1 = 100\ \text{g}$
• Molar mass of solvent (acetone), $M_1 = 58\ \text{g mol}^{-1}$
• Want: Molar mass of solute, $M_2 = ?$
Step 2: Note the Relevant Formula
To determine the molar mass of the solute from the lowering in vapour pressure, we can use the following relationship (from Raoult’s law):
$ \frac{P^\circ - P_s}{P_s} \;=\; \frac{w_2 \, M_1}{w_1 \, M_2} \,.$
Step 3: Substitute the Known Values into the Formula
• $P^\circ - P_s = 185 - 183 = 2\ \text{torr}$
• Hence,
$ \frac{2}{183} \;=\; \frac{1.2 \times 58}{100 \times M_2} \,. $
Step 4: Solve for the Molar Mass of the Solute ($M_2$)
First, compute the numerator on the right-hand side:
$1.2 \times 58 = 69.6.$
So the equation becomes:
$ \frac{2}{183} \;=\; \frac{69.6}{100 \times M_2}.$
Rewriting,
$ \frac{2}{183} \times 100 \times M_2 \;=\; 69.6.$
$ M_2 \;=\; \frac{69.6 \times 183}{2 \times 100} \,.$
Performing the division and multiplication gives approximately:
$ M_2 \approx 64\ \text{g mol}^{-1}.$
Step 5: Conclude the Molar Mass of the Solute
Therefore, the molar mass of the non-volatile substance is $64\ \text{g mol}^{-1}$.
