If 8 g of a non-electrolyte solute is dissolved in 114 g of n-octane to reduce its vapor pressure to 80%, the

Question

If 8 g of a non-electrolyte solute is dissolved in 114 g of n-octane to reduce its vapor pressure to 80%, the molar mass (in g mol⁻¹) of the solute is: [Molar mass of n-octane is 114 g mol⁻¹]

Accepted Answer

According to Raoult's law, the relative lowering of vapour pressure is equal to the mole fraction of the solute.

**Given:**
*   Mass of solute ($w_2$) = 8 g
*   Mass of solvent ($w_1$) = 114 g
*   Molar mass of solvent ($M_1$, n-octane) = 114 g mol⁻¹
*   Vapour pressure is reduced *to* 80%, so the relative lowering is ($100 - 80$)/100 = 0.20.

**Formula:**
For dilute solutions, the approximation $\frac{p_1^0 - p_1}{p_1^0} \approx \frac{n_2}{n_1}$ is often used .
$$\frac{p_1^0 - p_1}{p_1^0} = \frac{w_2 	imes M_1}{M_2 	imes w_1}$$

**Calculation:**
1.  **Calculate moles of solvent ($n_1$):** $n_1 = 114 	ext{ g} / 114 	ext{ g mol}^{-1} = 1 	ext{ mol}$.
2.  **Apply the formula:**
    $$ 0.20 = \frac{8 	ext{ g} 	imes 114 	ext{ g mol}^{-1}}{M_2 	imes 114 	ext{ g}} $$
    $$ 0.20 = \frac{8}{M_2} $$
    $$ M_2 = \frac{8}{0.20} = 40 	ext{ g mol}^{-1} $$

(Note: The approximation is used here to match the options provided in the examination).

Step-by-Step Solution

Practice All CHEMISTRY Questions