How many grams of sucrose must be added to `360g` of water to lower the vapour pressure by `1.19mm Hg` at a temperatue at which vapour pressure of pure water is `25 mmHg` ?

Options

A. 342 g
B. 360 g
C. 375 g
D. 380 g

Answer

To solve the problem of how many grams of sucrose must be added to 360 g of water to lower the vapor pressure by 1.19 mm Hg, we can follow these steps:

### Step 1: Understand Raoult's Law
Raoult's Law states that the relative lowering of vapor pressure of a solvent is equal to the mole fraction of the solute in the solution. The formula is given by:

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

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

### Step 2: Calculate the Relative Lowering of Vapor Pressure
Given:
- \(P_0 = 25 \, \text{mm Hg}\)
- Lowering of vapor pressure = \(1.19 \, \text{mm Hg}\)

Thus, the vapor pressure of the solution \(P\) is:

\[ P = P_0 - \text{lowering} = 25 \, \text{mm Hg} - 1.19 \, \text{mm Hg} = 23.81 \, \text{mm Hg} \]

Now, we can calculate the relative lowering of vapor pressure:

\[ \frac{P_0 - P}{P_0} = \frac{1.19}{25} = 0.0476 \]

### Step 3: Set Up the Mole Fraction Equation
Let \(n_{\text{solute}}\) be the number of moles of sucrose and \(n_{\text{solvent}}\) be the number of moles of water.

The mole fraction of the solute can be expressed as:

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

### Step 4: Calculate Moles of Solvent
To find \(n_{\text{solvent}}\), we can use the mass of water:

\[ n_{\text{solvent}} = \frac{\text{mass of water}}{\text{molar mass of water}} = \frac{360 \, \text{g}}{18 \, \text{g/mol}} = 20 \, \text{mol} \]

### Step 5: Substitute into the Mole Fraction Equation
Now, substituting the values into the mole fraction equation:

\[ 0.0476 = \frac{n_{\text{solute}}}{n_{\text{solute}} + 20} \]

### Step 6: Solve for Moles of Solute
Cross-multiplying gives:

\[ 0.0476(n_{\text{solute}} + 20) = n_{\text{solute}} \]

Expanding and rearranging:

\[ 0.0476n_{\text{solute}} + 0.952 = n_{\text{solute}} \]

\[ n_{\text{solute}} - 0.0476n_{\text{solute}} = 0.952 \]

\[ 0.952 = 0.9524n_{\text{solute}} \]

\[ n_{\text{solute}} = \frac{0.952}{0.9524} \approx 1 \, \text{mol} \]

### Step 7: Calculate the Mass of Sucrose
The molar mass of sucrose (C₁₂H₂₂O₁₁) is calculated as follows:

\[ \text{Molar mass of sucrose} = (12 \times 12) + (22 \times 1) + (11 \times 16) = 144 + 22 + 176 = 342 \, \text{g/mol} \]

Now, using the number of moles to find the mass:

\[ \text{mass of sucrose} = n_{\text{solute}} \times \text{molar mass of sucrose} = 1 \, \text{mol} \times 342 \, \text{g/mol} = 342 \, \text{g} \]

### Final Answer
Thus, the mass of sucrose required is **342 grams**.

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