The total vapour pressure of a 4 mole % solution of `NH_(3)` in water at 293 K is 50.0 torr. The vapour pressure of pure water is 17.0 torr at this temperature . Applying Henry's and Raoult's law, calculate the total vapour pressure for a 5 mole % solution:

Options

A. 58.25 torr
B. 33 torr
C. 42.1 torr
D. 52.25 torr

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Information
We are given:
- Total vapor pressure of a 4 mole % solution of NH₃ in water at 293 K = 50.0 torr
- Vapor pressure of pure water at 293 K = 17.0 torr

### Step 2: Calculate the Mole Fractions for the 4 mole % Solution
The mole fraction of ammonia (NH₃) in the 4 mole % solution can be calculated as:
- Mole fraction of NH₃ (X_NH₃) = 4/100 = 0.04
- Mole fraction of water (X_H₂O) = 1 - X_NH₃ = 1 - 0.04 = 0.96

### Step 3: Apply Raoult's Law
According to Raoult's Law, the total vapor pressure (P_total) can be expressed as:
\[ P_{\text{total}} = (X_{\text{NH}_3} \cdot P^0_{\text{NH}_3}) + (X_{\text{H}_2O} \cdot P^0_{\text{H}_2O}) \]
Where:
- \( P^0_{\text{H}_2O} = 17.0 \, \text{torr} \)
- \( P_{\text{total}} = 50.0 \, \text{torr} \)

Substituting the known values:
\[ 50.0 = (0.04 \cdot P^0_{\text{NH}_3}) + (0.96 \cdot 17.0) \]

### Step 4: Calculate the Contribution from Water
Calculate the contribution from water:
\[ 0.96 \cdot 17.0 = 16.32 \, \text{torr} \]

### Step 5: Solve for \( P^0_{\text{NH}_3} \)
Now substitute back into the equation:
\[ 50.0 = (0.04 \cdot P^0_{\text{NH}_3}) + 16.32 \]
Rearranging gives:
\[ 50.0 - 16.32 = 0.04 \cdot P^0_{\text{NH}_3} \]
\[ 33.68 = 0.04 \cdot P^0_{\text{NH}_3} \]

Now, solve for \( P^0_{\text{NH}_3} \):
\[ P^0_{\text{NH}_3} = \frac{33.68}{0.04} = 842 \, \text{torr} \]

### Step 6: Calculate the Mole Fractions for the 5 mole % Solution
For a 5 mole % solution:
- Mole fraction of NH₃ (X_NH₃) = 5/100 = 0.05
- Mole fraction of water (X_H₂O) = 1 - X_NH₃ = 1 - 0.05 = 0.95

### Step 7: Calculate the Total Vapor Pressure for the 5 mole % Solution
Using Raoult's Law again:
\[ P'_{\text{total}} = (X_{\text{NH}_3} \cdot P^0_{\text{NH}_3}) + (X_{\text{H}_2O} \cdot P^0_{\text{H}_2O}) \]
Substituting the values:
\[ P'_{\text{total}} = (0.05 \cdot 842) + (0.95 \cdot 17) \]

Calculating each part:
- Contribution from NH₃: \( 0.05 \cdot 842 = 42.1 \, \text{torr} \)
- Contribution from water: \( 0.95 \cdot 17 = 16.15 \, \text{torr} \)

### Step 8: Add the Contributions
Now add the contributions:
\[ P'_{\text{total}} = 42.1 + 16.15 = 58.25 \, \text{torr} \]

### Final Answer
The total vapor pressure for a 5 mole % solution of ammonia in water is **58.25 torr**.

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