How many grams of a non volatile solute having a molecular mass of 90 g/mole are to be dissolved in 97.5 g water in order to obtain relative lowering in the vapour pressure of 2.5 percent:

Options

A. 25
B. 18
C. 12.5
D. 9

Answer

To solve the problem of how many grams of a non-volatile solute with a molecular mass of 90 g/mole are to be dissolved in 97.5 g of water to achieve a relative lowering of vapor pressure of 2.5%, we can follow these steps:

### Step 1: Understand the Concept of Relative Lowering of Vapor Pressure
The relative lowering of vapor pressure (ΔP/P₀) is given by the formula:
\[ \frac{P_0 - P_s}{P_0} = \frac{n_2}{n_1 + n_2} \]
where \(P_0\) is the vapor pressure of the pure solvent, \(P_s\) is the vapor pressure of the solution, \(n_1\) is the number of moles of the solvent, and \(n_2\) is the number of moles of solute.

### Step 2: Set Up the Given Information
- Relative lowering of vapor pressure = 2.5% = 0.025
- Mass of water (solvent) = 97.5 g
- Molecular mass of water (H₂O) = 18 g/mole
- Molecular mass of solute = 90 g/mole

### Step 3: Calculate the Moles of the Solvent (Water)
Using the formula:
\[ n_1 = \frac{\text{mass of solvent}}{\text{molar mass of solvent}} = \frac{97.5 \text{ g}}{18 \text{ g/mole}} = 5.4167 \text{ moles} \]

### Step 4: Relate the Relative Lowering of Vapor Pressure to Moles of Solute
From the relative lowering of vapor pressure:
\[ \frac{n_2}{n_1 + n_2} = 0.025 \]
Let \(n_2\) be the number of moles of solute. Rearranging gives:
\[ n_2 = 0.025(n_1 + n_2) \]
This can be simplified to:
\[ n_2 - 0.025 n_2 = 0.025 n_1 \]
\[ 0.975 n_2 = 0.025 n_1 \]
\[ n_2 = \frac{0.025 n_1}{0.975} \]

### Step 5: Substitute the Value of \(n_1\)
Substituting \(n_1 = 5.4167\):
\[ n_2 = \frac{0.025 \times 5.4167}{0.975} \approx 0.1392 \text{ moles} \]

### Step 6: Calculate the Mass of the Solute
Using the number of moles of solute to find the mass:
\[ \text{mass of solute} = n_2 \times \text{molar mass of solute} = 0.1392 \text{ moles} \times 90 \text{ g/mole} \approx 12.528 \text{ g} \]

### Step 7: Round the Answer
Thus, the mass of the non-volatile solute to be dissolved is approximately:
\[ \text{mass of solute} \approx 12.5 \text{ g} \]

### Final Answer
The mass of the non-volatile solute required is **12.5 grams**.

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