Raoult's Law

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The vapor pressure of pure benzene at 298 K is 100 mm Hg. When 2.0 g of a non-volatile solute is dissolved in 78 g of benzene, the vapor pressure drops to 98 mm Hg. The molar mass of the solute is:

RankGuru · Accepted Answer

According to **Raoult's Law** for relative lowering of vapor pressure:
$$\frac{P^0 - P_s}{P^0} = X_{solute} \approx \frac{n_2}{n_1}$$
- **Values**: $P^0 = 100 	ext{ mm Hg}$, $P_s = 98 	ext{ mm Hg}$.
- Moles of benzene ($n_1$) = $\frac{	ext{mass}}{	ext{molar mass}} = \frac{78}{78} = 1 	ext{ mol}$.
- Fractional lowering: $\frac{100 - 98}{100} = 0.02$.
- Since $X_{solute} = 0.02$ and $n_1 = 1$, we have $\frac{n_2}{n_1} \approx 0.02 \Rightarrow n_2 = 0.02 	ext{ mol}$.
- Molar mass $M_2 = \frac{	ext{mass of solute}}{n_2} = \frac{2.0}{0.02} = 100 	ext{ g/mol}$.

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