Calculate the weight of non`-` volatile solute having molecular weight 40, which should be dissolvd in `57 gm` octane to reduce its vapour pressure to `80% :`

Options

A. 47.2 g
B. 5 g
C. 106.2 g
D. None of these

Answer

To solve the problem of calculating the weight of a non-volatile solute that must be dissolved in 57 grams of octane to reduce its vapor pressure to 80%, we can follow these steps:

### Step 1: Understand the Given Data
- Weight of the solvent (octane) = 57 g
- Molecular weight of the solute = 40 g/mol
- Vapor pressure reduction to 80% means the vapor pressure of the solution is 80% of the pure solvent's vapor pressure.

### Step 2: Define the Variables
Let:
- \( P_0 \) = vapor pressure of pure octane
- \( P_s \) = vapor pressure of the solution
- \( W \) = weight of the solute to be calculated

Since the vapor pressure is reduced to 80%, we can express this as:
\[ P_s = 0.8 P_0 \]

### Step 3: Use Raoult's Law
According to Raoult's Law, the change in vapor pressure can be expressed as:
\[ \frac{P_0 - P_s}{P_s} = \frac{n_{solute}}{n_{solvent}} \]
Where:
- \( n_{solute} \) = moles of solute
- \( n_{solvent} \) = moles of solvent

### Step 4: Calculate the Moles of Solvent
The moles of octane (solvent) can be calculated using its molar mass:
\[ \text{Molar mass of octane} = 114 \text{ g/mol} \]
\[ n_{solvent} = \frac{57 \text{ g}}{114 \text{ g/mol}} = 0.5 \text{ mol} \]

### Step 5: Substitute Values into Raoult's Law
Substituting \( P_s \) into the equation:
\[ \frac{P_0 - 0.8 P_0}{0.8 P_0} = \frac{n_{solute}}{0.5} \]
This simplifies to:
\[ \frac{0.2 P_0}{0.8 P_0} = \frac{n_{solute}}{0.5} \]
\[ \frac{0.2}{0.8} = \frac{n_{solute}}{0.5} \]
\[ 0.25 = \frac{n_{solute}}{0.5} \]

### Step 6: Calculate Moles of Solute
From the equation above, we can find \( n_{solute} \):
\[ n_{solute} = 0.25 \times 0.5 = 0.125 \text{ mol} \]

### Step 7: Calculate the Weight of the Solute
Using the molecular weight of the solute:
\[ W = n_{solute} \times \text{Molar mass of solute} \]
\[ W = 0.125 \text{ mol} \times 40 \text{ g/mol} = 5 \text{ g} \]

### Conclusion
The weight of the non-volatile solute that should be dissolved in 57 grams of octane to reduce its vapor pressure to 80% is **5 grams**.