Question

What is vapour pressure of a solution containing 0.1 mol of non-volatile solute dissolved in 16.2 g of water ? ($P_1^0= 32\text{ mm Hg}$)

• A. $21.6\text{ mm Hg}$
• B. $28.8\text{ mm Hg}$
• C. $15.7\text{ mm Hg}$
• D. $18.1\text{ mm Hg}$

Hint:

Always ensure you are calculating the mole fraction of the solvent ($x_1$) when using $P = P^0 \cdot x_1$, not the solute! If you calculate the mole fraction of the solute ($x_2$), use relative lowering: $\frac{P^0 - P}{P^0} = x_2$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to calculate the vapor pressure of a solution formed by adding a known amount of non-volatile solute to a known mass of solvent (water).

Step 2: Key Formula or Approach:
According to Raoult's Law for a solution containing a non-volatile solute, the vapor pressure of the solution ($P_1$) is directly proportional to the mole fraction of the solvent ($x_1$).
$$P_1 = P_1^0 \times x_1$$
Where $P_1^0$ is the vapor pressure of the pure solvent, and $x_1 = \frac{n_1}{n_1 + n_2}$.

Step 3: Detailed Explanation:
First, calculate the number of moles of the solute ($n_2$) and the solvent ($n_1$).
$n_2 = 0.1\text{ mol}$ (given)
The molar mass of water ($\text{H}_2\text{O}$) is $18\text{ g/mol}$.
$n_1 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{16.2\text{ g}}{18\text{ g/mol}} = 0.9\text{ mol}$
Next, calculate the mole fraction of the solvent ($x_1$):
$$x_1 = \frac{n_1}{n_1 + n_2} = \frac{0.9}{0.9 + 0.1} = \frac{0.9}{1.0} = 0.9$$
Now, substitute this into Raoult's Law equation:
$$P_1 = 32\text{ mm Hg} \times 0.9$$
$$P_1 = 28.8\text{ mm Hg}$$

Step 4: Final Answer:
The vapor pressure of the solution is $28.8\text{ mm Hg}$, matching option (B).