Calculate mole fraction, molarity and molality of `C_(2)H_(5)OH` solution which is `50%` by weight of `C_(2)H_(5)OH` in `H_(2)O` and has a density of `.9144` g per "cc".

Answer

To solve the problem of calculating the mole fraction, molarity, and molality of a 50% by weight solution of ethyl alcohol (C₂H₅OH) in water (H₂O) with a density of 0.9144 g/cc, we can follow these steps:

### Step 1: Calculate the mass of the solution
Assume we have 100 g of the solution. Since the solution is 50% by weight of C₂H₅OH, we have:
- Mass of C₂H₅OH = 50 g
- Mass of H₂O = 50 g

### Step 2: Calculate the number of moles of C₂H₅OH
The molar mass of C₂H₅OH (ethyl alcohol) is approximately 46.07 g/mol. Therefore, the number of moles of C₂H₅OH can be calculated as follows:
\[ \text{Number of moles of C₂H₅OH} = \frac{\text{mass}}{\text{molar mass}} = \frac{50 \, \text{g}}{46.07 \, \text{g/mol}} \approx 1.085 \, \text{mol} \]

### Step 3: Calculate the number of moles of H₂O
The molar mass of H₂O is approximately 18.02 g/mol. Therefore, the number of moles of H₂O can be calculated as follows:
\[ \text{Number of moles of H₂O} = \frac{50 \, \text{g}}{18.02 \, \text{g/mol}} \approx 2.775 \, \text{mol} \]

### Step 4: Calculate the mole fraction of C₂H₅OH
The mole fraction \(X\) of C₂H₅OH is calculated using the formula:
\[ X_{\text{C₂H₅OH}} = \frac{\text{moles of C₂H₅OH}}{\text{moles of C₂H₅OH} + \text{moles of H₂O}} = \frac{1.085}{1.085 + 2.775} \approx 0.281 \]

### Step 5: Calculate the volume of the solution
Using the density of the solution, we can find the volume of the 100 g solution:
\[ \text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{100 \, \text{g}}{0.9144 \, \text{g/cc}} \approx 109.36 \, \text{cc} \approx 0.10936 \, \text{L} \]

### Step 6: Calculate the molarity of C₂H₅OH
Molarity (M) is defined as the number of moles of solute per liter of solution:
\[ \text{Molarity} = \frac{\text{moles of C₂H₅OH}}{\text{volume of solution in L}} = \frac{1.085}{0.10936} \approx 9.90 \, \text{M} \]

### Step 7: Calculate the molality of C₂H₅OH
Molality (m) is defined as the number of moles of solute per kilogram of solvent. The mass of the solvent (H₂O) is 50 g or 0.050 kg:
\[ \text{Molality} = \frac{\text{moles of C₂H₅OH}}{\text{mass of solvent in kg}} = \frac{1.085}{0.050} \approx 21.70 \, \text{m} \]

### Summary of Results
- Mole fraction of C₂H₅OH: \(X_{\text{C₂H₅OH}} \approx 0.281\)
- Molarity of C₂H₅OH: \( \approx 9.90 \, \text{M} \)
- Molality of C₂H₅OH: \( \approx 21.70 \, \text{m} \)