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The vapour pressure of ethyl alcohol and...

The vapour pressure of ethyl alcohol and methyl alcohol are 45mm and 90 mm. An ideal solution is formed at the same temperature by mixing 60g of `C_(2)H_(5) OH ` with 40g of `CH_(3)OH`. Total vapour pressure of the solution is approximately.

A

67 mm

B

35 mm

C

105 mm

D

140 mm

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To find the total vapor pressure of the solution formed by mixing ethyl alcohol (C2H5OH) and methyl alcohol (CH3OH), we can use Raoult's Law, which states that the vapor pressure of a solution is equal to the sum of the partial pressures of each component. ### Step-by-Step Solution: 1. **Determine the Vapor Pressures of Pure Components**: - Vapor pressure of pure ethyl alcohol (P0E) = 45 mm Hg - Vapor pressure of pure methyl alcohol (P0M) = 90 mm Hg 2. **Calculate the Molar Masses**: - Molar mass of ethyl alcohol (C2H5OH): - C: 2 × 12 g/mol = 24 g/mol - H: 6 × 1 g/mol = 6 g/mol - O: 1 × 16 g/mol = 16 g/mol - Total = 24 + 6 + 16 = 46 g/mol - Molar mass of methyl alcohol (CH3OH): - C: 1 × 12 g/mol = 12 g/mol - H: 4 × 1 g/mol = 4 g/mol - O: 1 × 16 g/mol = 16 g/mol - Total = 12 + 4 + 16 = 32 g/mol 3. **Calculate the Number of Moles**: - Number of moles of ethyl alcohol: \[ n_E = \frac{60 \text{ g}}{46 \text{ g/mol}} \approx 1.304 \text{ moles} \] - Number of moles of methyl alcohol: \[ n_M = \frac{40 \text{ g}}{32 \text{ g/mol}} \approx 1.25 \text{ moles} \] 4. **Calculate the Mole Fractions**: - Total number of moles: \[ n_{total} = n_E + n_M = 1.304 + 1.25 \approx 2.554 \text{ moles} \] - Mole fraction of ethyl alcohol (XE): \[ X_E = \frac{n_E}{n_{total}} = \frac{1.304}{2.554} \approx 0.510 \] - Mole fraction of methyl alcohol (XM): \[ X_M = \frac{n_M}{n_{total}} = \frac{1.25}{2.554} \approx 0.490 \] 5. **Calculate the Total Vapor Pressure**: - Using Raoult's Law: \[ P_{total} = P0_E \cdot X_E + P0_M \cdot X_M \] - Substitute the values: \[ P_{total} = (45 \text{ mm Hg} \cdot 0.510) + (90 \text{ mm Hg} \cdot 0.490) \] \[ P_{total} = 22.95 + 44.1 = 67.05 \text{ mm Hg} \] 6. **Final Answer**: - The total vapor pressure of the solution is approximately **67 mm Hg**.

To find the total vapor pressure of the solution formed by mixing ethyl alcohol (C2H5OH) and methyl alcohol (CH3OH), we can use Raoult's Law, which states that the vapor pressure of a solution is equal to the sum of the partial pressures of each component. ### Step-by-Step Solution: 1. **Determine the Vapor Pressures of Pure Components**: - Vapor pressure of pure ethyl alcohol (P0E) = 45 mm Hg - Vapor pressure of pure methyl alcohol (P0M) = 90 mm Hg ...
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