Methanol and ethanol form nearly ideal solution at 300 K. A solution is made by mixing 32g methanol and 23g ethanol. Calculate the partial pressure of its constituents and total pressure of the solution. `["at "300 K.p^(0)(CH_(3)OH)=90 mm Hg, p(C_(2)H_(5)OH)=51 mm Hg]`

Explanation

To solve the problem of calculating the partial pressures of methanol and ethanol in a nearly ideal solution, as well as the total pressure of the solution, we can follow these steps:

### Step 1: Calculate the number of moles of methanol (CH₃OH) and ethanol (C₂H₅OH).

1. **Molecular mass of methanol (CH₃OH)** = 32 g/mol
2. **Molecular mass of ethanol (C₂H₅OH)** = 46 g/mol

Using the formula for moles:
\[ \text{Moles} = \frac{\text{mass}}{\text{molecular mass}} \]

- For methanol:
\[ \text{Moles of CH₃OH} = \frac{32 \, \text{g}}{32 \, \text{g/mol}} = 1 \, \text{mol} \]

- For ethanol:
\[ \text{Moles of C₂H₅OH} = \frac{23 \, \text{g}}{46 \, \text{g/mol}} = 0.5 \, \text{mol} \]

### Step 2: Calculate the total number of moles in the solution.

\[ \text{Total moles} = \text{Moles of CH₃OH} + \text{Moles of C₂H₅OH} = 1 + 0.5 = 1.5 \, \text{mol} \]

### Step 3: Calculate the mole fractions of methanol and ethanol.

- Mole fraction of methanol (X₁):
\[ X_{\text{CH₃OH}} = \frac{\text{Moles of CH₃OH}}{\text{Total moles}} = \frac{1}{1.5} = 0.67 \]

- Mole fraction of ethanol (X₂):
\[ X_{\text{C₂H₅OH}} = \frac{\text{Moles of C₂H₅OH}}{\text{Total moles}} = \frac{0.5}{1.5} = 0.33 \]

### Step 4: Calculate the partial pressures of methanol and ethanol.

Using Raoult's Law:
\[ P_i = P_i^0 \cdot X_i \]

Where \(P_i^0\) is the vapor pressure of the pure component and \(X_i\) is the mole fraction.

- For methanol:
\[ P_{\text{CH₃OH}} = P_{\text{CH₃OH}}^0 \cdot X_{\text{CH₃OH}} = 90 \, \text{mm Hg} \cdot 0.67 = 60.3 \, \text{mm Hg} \]

- For ethanol:
\[ P_{\text{C₂H₅OH}} = P_{\text{C₂H₅OH}}^0 \cdot X_{\text{C₂H₅OH}} = 51 \, \text{mm Hg} \cdot 0.33 = 16.83 \, \text{mm Hg} \]

### Step 5: Calculate the total pressure of the solution.

\[ P_{\text{total}} = P_{\text{CH₃OH}} + P_{\text{C₂H₅OH}} = 60.3 \, \text{mm Hg} + 16.83 \, \text{mm Hg} = 77.13 \, \text{mm Hg} \]

### Final Results:
- Partial pressure of methanol: **60.3 mm Hg**
- Partial pressure of ethanol: **16.83 mm Hg**
- Total pressure of the solution: **77.13 mm Hg**

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