The molality of a solution of ethyl alcohol `(C_(2)H_(5)OH)` in water is 1.55 m . How many grams of ethyl alcohol are dissolved in 2 kg of water ?

Explanation

To solve the problem, we need to find out how many grams of ethyl alcohol (C₂H₅OH) are dissolved in 2 kg of water, given that the molality of the solution is 1.55 m.

### Step-by-Step Solution:

1. **Understand the Definition of Molality**:
Molality (m) is defined as the number of moles of solute per kilogram of solvent. The formula is:
\[ \text{Molality} (m) = \frac{\text{Number of moles of solute}}{\text{Weight of solvent (kg)}} \]

2. **Set Up the Equation**:
We know the molality of the solution is 1.55 m, and we have 2 kg of water as the solvent. Let \( n \) be the number of moles of ethyl alcohol. Therefore, we can write:
\[ 1.55 = \frac{n}{2} \]

3. **Calculate the Number of Moles of Ethyl Alcohol**:
Rearranging the equation to find \( n \):
\[ n = 1.55 \times 2 = 3.1 \text{ moles} \]

4. **Determine the Molecular Weight of Ethyl Alcohol**:
The molecular formula of ethyl alcohol is C₂H₅OH. To find its molecular weight, we calculate:
- Carbon (C): 2 atoms × 12 g/mol = 24 g/mol
- Hydrogen (H): 6 atoms × 1 g/mol = 6 g/mol
- Oxygen (O): 1 atom × 16 g/mol = 16 g/mol
Adding these together:
\[ \text{Molecular weight of C₂H₅OH} = 24 + 6 + 16 = 46 \text{ g/mol} \]

5. **Calculate the Weight of Ethyl Alcohol**:
Now, we can find the weight of ethyl alcohol using the number of moles and its molecular weight:
\[ \text{Weight} = \text{Number of moles} \times \text{Molecular weight} \]
\[ \text{Weight} = 3.1 \text{ moles} \times 46 \text{ g/mol} = 142.6 \text{ grams} \]

### Final Answer:
The weight of ethyl alcohol dissolved in 2 kg of water is **142.6 grams**.