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. 66.67 g
B. 120 g
C. 30 g
D. 24 g

Answer

To solve the problem, we need to determine the amount of solute required to lower the vapor pressure of water by 10%. We will follow these steps:

### Step 1: Understand the given data
- Molar mass of solute = 60 g/mol
- Mass of water (solvent) = 180 g
- Vapor pressure lowering = 10%

### Step 2: Calculate the moles of water (solvent)
The number of moles of water can be calculated using the formula:
\[ \text{Moles of water} (n_1) = \frac{\text{mass of water}}{\text{molar mass of water}} \]
The molar mass of water (H₂O) is:
\[ \text{Molar mass of H₂O} = 2 \times 1 + 16 = 18 \text{ g/mol} \]
Now, substituting the values:
\[ n_1 = \frac{180 \text{ g}}{18 \text{ g/mol}} = 10 \text{ moles} \]

### Step 3: Calculate the change in vapor pressure
The vapor pressure lowering is given as 10%, which means:
\[ \Delta P = 0.1 \times P_0 \]
Where \(P_0\) is the original vapor pressure of water.

### Step 4: Use Raoult's Law
According to Raoult's Law, the relative lowering of vapor pressure is given by:
\[ \frac{\Delta P}{P_0} = \frac{n_2}{n_1 + n_2} \]
Where:
- \(n_2\) = moles of solute
- \(n_1\) = moles of solvent (water)

From the previous step, we know \(n_1 = 10\) moles. The relative lowering of vapor pressure is 0.1 (10%).

Substituting the values into the equation:
\[ 0.1 = \frac{n_2}{10 + n_2} \]

### Step 5: Solve for moles of solute (n2)
Cross-multiplying gives:
\[ 0.1(10 + n_2) = n_2 \]
Expanding this:
\[ 1 + 0.1n_2 = n_2 \]
Rearranging the equation:
\[ 1 = n_2 - 0.1n_2 \]
\[ 1 = 0.9n_2 \]
\[ n_2 = \frac{1}{0.9} \approx 1.111 \text{ moles} \]

### Step 6: Calculate the mass of solute (W)
Now we can find the mass of the solute using the formula:
\[ \text{mass of solute} (W) = n_2 \times \text{molar mass of solute} \]
Substituting the values:
\[ W = 1.111 \text{ moles} \times 60 \text{ g/mol} \approx 66.67 \text{ g} \]

### Conclusion
The amount of solute that must be added to 180 g of water to lower the vapor pressure by 10% is approximately **66.67 g**.

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