The mass of a non-volatile solute of molar mass `40g" "mol^(-1)` that should be dissolved in 114 g of octane to lower its vapour pressure by 20% is

Options

A. 11.4 g
B. 9.8 g
C. 12.8 g
D. 10g

Answer

To solve the problem of finding the mass of a non-volatile solute that should be dissolved in 114 g of octane to lower its vapor pressure by 20%, we can follow these steps:

### Step 1: Understand the Problem
We know that the vapor pressure of the solution is reduced by 20%. This means that if the vapor pressure of pure octane is 100%, the vapor pressure of the solution becomes 80%.

### Step 2: Use Raoult's Law
According to Raoult's Law, the relative lowering of vapor pressure is given by:
\[ \frac{P_0 - P}{P_0} = X_{\text{solute}} \]
Where:
- \(P_0\) = vapor pressure of pure solvent (octane)
- \(P\) = vapor pressure of the solution
- \(X_{\text{solute}}\) = mole fraction of the solute

Given that the vapor pressure of the solution is 80% of the pure solvent, we can express this as:
\[ \frac{100 - 80}{100} = 0.2 = X_{\text{solute}} \]

### Step 3: Calculate Moles of Solvent
Next, we need to calculate the number of moles of octane (the solvent). The molar mass of octane (C8H18) is calculated as follows:
\[ \text{Molar mass of octane} = (12 \times 8) + (1 \times 18) = 96 + 18 = 114 \, \text{g/mol} \]
Now, we can find the number of moles of octane:
\[ \text{Number of moles of octane} = \frac{\text{mass of octane}}{\text{molar mass of octane}} = \frac{114 \, \text{g}}{114 \, \text{g/mol}} = 1 \, \text{mol} \]

### Step 4: Set Up the Mole Fraction Equation
Let \(m\) be the mass of the solute. The number of moles of the solute can be expressed as:
\[ \text{Number of moles of solute} = \frac{m}{40} \, \text{mol} \]
Now, we can set up the equation for the mole fraction of the solute:
\[ X_{\text{solute}} = \frac{\text{moles of solute}}{\text{moles of solute} + \text{moles of solvent}} = \frac{\frac{m}{40}}{\frac{m}{40} + 1} \]
Setting this equal to the mole fraction we calculated earlier:
\[ 0.2 = \frac{\frac{m}{40}}{\frac{m}{40} + 1} \]

### Step 5: Solve for Mass of Solute
Cross-multiplying gives:
\[ 0.2 \left(\frac{m}{40} + 1\right) = \frac{m}{40} \]
Expanding this:
\[ 0.2 \cdot \frac{m}{40} + 0.2 = \frac{m}{40} \]
Rearranging gives:
\[ 0.2 = \frac{m}{40} - 0.2 \cdot \frac{m}{40} \]
\[ 0.2 = \frac{m}{40} (1 - 0.2) = \frac{m}{40} \cdot 0.8 \]
Now, solving for \(m\):
\[ 0.2 = \frac{0.8m}{40} \]
\[ m = \frac{0.2 \times 40}{0.8} = \frac{8}{0.8} = 10 \, \text{g} \]

### Final Answer
The mass of the non-volatile solute that should be dissolved is **10 grams**.

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