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 must 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
Relative lowering of vapor pressure is defined as:
\[ \text{Relative lowering} = \frac{P_0 - P}{P_0} = \text{mole fraction of solute} \]
Where \(P_0\) is the vapor pressure of the pure solvent and \(P\) is the vapor pressure of the solution.

### Step 2: Relate Relative Lowering to Mole Fraction
For a dilute solution, the mole fraction of the solute can be approximated as:
\[ \text{mole fraction of solute} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}} \approx \frac{n_{\text{solute}}}{n_{\text{solvent}}} \]
Where \(n_{\text{solute}}\) is the number of moles of the solute and \(n_{\text{solvent}}\) is the number of moles of the solvent.

### Step 3: Calculate Moles of Solvent
The number of moles of water (solvent) can be calculated using its mass and molecular weight:
\[ n_{\text{solvent}} = \frac{\text{mass of solvent}}{\text{molecular weight of solvent}} = \frac{97.5 \text{ g}}{18 \text{ g/mole}} = 5.4167 \text{ moles} \]

### Step 4: Set Up the Equation for Relative Lowering
Given that the relative lowering of vapor pressure is 2.5%, we can express this as:
\[ 0.025 = \frac{n_{\text{solute}}}{n_{\text{solvent}}} \]
Substituting the value of \(n_{\text{solvent}}\):
\[ 0.025 = \frac{n_{\text{solute}}}{5.4167} \]

### Step 5: Solve for Moles of Solute
Rearranging the equation gives:
\[ n_{\text{solute}} = 0.025 \times 5.4167 = 0.1354 \text{ moles} \]

### Step 6: Convert Moles of Solute to Grams
Now, we can convert the number of moles of solute to grams using its molecular weight:
\[ \text{mass of solute} = n_{\text{solute}} \times \text{molecular weight of solute} = 0.1354 \text{ moles} \times 90 \text{ g/mole} = 12.054 \text{ g} \]

### Step 7: Round the Answer
Rounding to an appropriate number of significant figures, we find:
\[ \text{mass of solute} \approx 12.1 \text{ g} \]

### Final Answer
Approximately **12.1 grams** of the non-volatile solute are needed.

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