Animated Solution for Chemistry - Solutions: The vapour pressure of acetone at $20^{\circ }\text{C}$ is $185\text{ torr}$. When $1.2\text{ g}$ of a non-volatile substance was dissolved in $100\text{ g}$ of acetone at $20^{\circ }\text{C}$, its vapour pressure was $183\text{ torr}$. The molar mass (${\text{g mol}}^{-1}$) of the substance is

Imagine you are standing in a chemistry lab, looking at a beaker filled with exactly $100\text{ g}$ of pure acetone. At a comfortable room temperature of $20^{\circ }\text{C}$, the acetone molecules are highly energetic. They constantly escape the liquid surface, creating a vapor pressure of $185\text{ torr}$.

Now, the plot thickens. We introduce a mystery guest: $1.2\text{ g}$ of an unknown, non-volatile substance. As these new particles dissolve and occupy space at the surface, they physically block some of the acetone molecules from escaping. Consequently, the vapor pressure drops to $183\text{ torr}$. Our mission is to use this pressure drop to deduce the molar mass of our mystery guest.

To connect the macroscopic drop in vapor pressure to the microscopic amount of solute, we turn to Raoult's Law. The standard textbook formula for the relative lowering of vapor pressure is:

While this is perfectly correct, dealing with the unknown moles ($n$) in the denominator can lead to messy algebra. Instead, we can use a brilliant mathematical manipulation. By inverting both sides and subtracting $1$, we arrive at a much more elegant and exact derived form:

Notice the subtle change? The denominator on the left is now $p_s$ (the solution's vapor pressure), and the right side is simply the ratio of moles. This exact formula is a powerful tool that saves time and prevents approximation errors!

Let's break down the right side of our master equation. We need expressions for the moles of both the solute ($n$) and the solvent ($N$).

For our unknown solute, the moles are simply its given mass divided by its unknown molar mass ($M$):

For our solvent, acetone (${\text{CH}}_3{\text{COCH}}_3$), we first need its molar mass. Calculating it gives $(3\times 12)+(6\times 1)+16=58\text{ g/mol}$. Therefore, the moles of acetone are:

Now, we carefully substitute all our known values into the master formula:

Let's simplify the equation. The numerator on the left is just $2$. On the right side, the fraction $\frac{100}{58}$ flips up, allowing the $58$ to multiply with the $1.2$, while the capital $M$ drops down to join the $100$:

It is time to isolate our target, $M$. By cross-multiplying, we bring $M$ to the left side:

Crunching these numbers gives us exactly $63.684{\text{ g mol}}^{-1}$. Looking at our multiple-choice options, we can confidently round this to the nearest integer.