A current of dry air was passed first through a series of bulbs containing a solution of `C_(6)H_(5) - NO_(2` in ethanol of molality 0.725 and then through a series of bulbs containing pure ethanol. (T = 284 K) loss in weight of the solvent bulbs was 0.0685 g. Calculate the loss in weight of the solution bulbs.

Options

A. 4.60 g
B. 5.20 g
C. 2.50 g
D. 2.05 g

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the given data
- Molality of the solution, \( m = 0.725 \)
- Loss in weight of the solvent bulbs, \( \Delta W_{\text{solvent}} = 0.0685 \, \text{g} \)
- Molar mass of ethanol, \( M_2 = 46 \, \text{g/mol} \)
- Temperature, \( T = 284 \, \text{K} \) (not directly needed for calculations)

### Step 2: Calculate the ratio of moles of solvent to moles of solute
Using the formula for molality:
\[ m = \frac{n_1}{n_2} \cdot \frac{1000}{M_2} \]
Where:
- \( n_1 \) = number of moles of solute (C₆H₅NO₂)
- \( n_2 \) = number of moles of solvent (ethanol)

Rearranging gives:
\[ \frac{n_2}{n_1} = \frac{m \cdot M_2}{1000} \]
Substituting the values:
\[ \frac{n_2}{n_1} = \frac{0.725 \cdot 46}{1000} = 0.03335 \]

### Step 3: Apply Raoult's Law
According to Raoult's Law:
\[ \frac{P_0 - P_s}{P_s} = \frac{n_2}{n_1} \]
Where:
- \( P_0 \) = vapor pressure of pure solvent
- \( P_s \) = vapor pressure of the solution

This can be rearranged to express the loss in weight of the solution bulbs:
\[ P_0 - P_s = P_s \cdot \frac{n_2}{n_1} \]
Given that the loss in weight of the solvent bulbs corresponds to \( P_0 - P_s \):
\[ 0.0685 = P_s \cdot 0.03335 \]

### Step 4: Solve for \( P_s \)
Rearranging the equation gives:
\[ P_s = \frac{0.0685}{0.03335} \approx 2.05 \, \text{g} \]

### Conclusion
The loss in weight of the solution bulbs is approximately **2.05 g**.

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