The amount of solute (molar mass 60 g `mol^(-1)`) that must be added to 180 g of water so that the vapour pressure of water is lowered by `10%` is

Options

A. 60 g
B. 120 g
C. 30 g
D. 24 g

Answer

To solve the problem, we need to determine the amount of solute that must be added to 180 g of water to lower the vapor pressure of water by 10%. Here’s a step-by-step solution:

### Step 1: Understand the relationship between vapor pressure and mole fraction
The formula for the lowering of vapor pressure is given by:

\[ \frac{P_0 - P_s}{P_0} = x_{\text{solute}} \]

Where:
- \( P_0 \) = vapor pressure of pure solvent (water)
- \( P_s \) = vapor pressure of the solution
- \( x_{\text{solute}} \) = mole fraction of the solute

### Step 2: Set up the equation
Given that the vapor pressure is lowered by 10%, we can express this as:

\[ P_0 - P_s = 0.10 P_0 \]

Assuming \( P_0 = 100 \) (for simplicity), we have:

\[ P_s = P_0 - 0.10 P_0 = 90 \]

Now substituting into the formula:

\[ \frac{100 - 90}{100} = x_{\text{solute}} \implies x_{\text{solute}} = 0.10 \]

### Step 3: Relate mole fraction to moles of solute and solvent
The mole fraction of the solute can be expressed as:

\[ x_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}} \]

Where:
- \( n_{\text{solute}} \) = moles of solute
- \( n_{\text{solvent}} \) = moles of solvent (water)

### Step 4: Calculate moles of solvent
The molar mass of water is 18 g/mol. Therefore, the number of moles of water is:

\[ n_{\text{solvent}} = \frac{180 \text{ g}}{18 \text{ g/mol}} = 10 \text{ moles} \]

### Step 5: Set up the equation for mole fraction
Now substituting \( x_{\text{solute}} = 0.10 \) into the mole fraction equation:

\[ 0.10 = \frac{n_{\text{solute}}}{n_{\text{solute}} + 10} \]

### Step 6: Solve for moles of solute
Let \( n_{\text{solute}} = x \). Then we have:

\[ 0.10 = \frac{x}{x + 10} \]

Cross-multiplying gives:

\[ 0.10(x + 10) = x \]

Expanding and rearranging:

\[ 0.10x + 1 = x \implies 1 = x - 0.10x \implies 1 = 0.90x \implies x = \frac{1}{0.90} \approx 1.111 \text{ moles} \]

### Step 7: Convert moles of solute to grams
Now, we need to convert moles of solute to grams. Given the molar mass of the solute is 60 g/mol:

\[ \text{Mass of solute} = n_{\text{solute}} \times \text{molar mass} = 1.111 \text{ moles} \times 60 \text{ g/mol} \approx 66.67 \text{ g} \]

### Step 8: Round to the nearest whole number
Since the question asks for the amount of solute, we can round this to:

\[ \text{Mass of solute} \approx 60 \text{ g} \]

Thus, the final answer is:

**60 g**