A current of dry air was passed first through a series of bulbs containing a solution of `C_(6)H_(5) - NO_(20` 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 need to calculate the loss in weight of the solution bulbs based on the given data. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Given Data
- The molality (m) of the solution is 0.725.
- The loss in weight of the solvent bulbs (pure ethanol) is 0.0685 g.
- The temperature (T) is 284 K, but it is not directly needed for this calculation.

### Step 2: Calculate the Molar Mass of Ethanol
The chemical formula of ethanol is C₂H₅OH. We can calculate its molar mass as follows:
- Molar mass of Carbon (C) = 12 g/mol (2 Carbons = 2 × 12 = 24 g/mol)
- Molar mass of Hydrogen (H) = 1 g/mol (6 Hydrogens = 6 × 1 = 6 g/mol)
- Molar mass of Oxygen (O) = 16 g/mol

Thus, the molar mass of ethanol (C₂H₅OH) = 24 + 6 + 16 = 46 g/mol.

### Step 3: Calculate the Number of Moles of Solvent (Ethanol)
Using the definition of molality:
\[ \text{molality (m)} = \frac{n_1}{n_2} \]
Where:
- \( n_1 \) = number of moles of solute (C₆H₅NO₂)
- \( n_2 \) = number of moles of solvent (ethanol)

From the formula, we can rearrange it to find the ratio of moles:
\[ \frac{n_2}{n_1} = m \times \frac{M_2}{1000} \]
Where \( M_2 \) is the molar mass of the solvent (ethanol).

Substituting the values:
\[ \frac{n_2}{n_1} = 0.725 \times \frac{46}{1000} \]
\[ \frac{n_2}{n_1} = 0.03335 \]

### Step 4: 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

We know that the loss in weight of the solvent bulbs is proportional to the difference in vapor pressures:
\[ P_0 - P_s \propto \text{loss in weight of solvent} \]
Thus:
\[ P_0 - P_s = k \cdot \text{loss in weight of solvent} \]
Where \( k \) is a proportionality constant.

Given that the loss in weight of the solvent bulbs is 0.0685 g, we can express this as:
\[ P_0 - P_s = 0.0685 \]

### Step 5: Calculate the Loss in Weight of the Solution Bulbs
Now substituting into the Raoult's Law equation:
\[ \frac{0.0685}{P_s} = 0.03335 \]
Rearranging gives:
\[ P_s = \frac{0.0685}{0.03335} \]
Calculating this gives:
\[ P_s \approx 2.05 \text{ g} \]

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