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3 years ago

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A solution is prepared by mixing 0.0300 mol CH2Cl2 and 0.0500 mol CH2Br2 at 25 degrees C. Assuming the solution is ideal, calcul

ate the composition of the vapor (in terms of mole fractions) at 25 degrees C. At 25 degrees C, the vapor pressures of pure CH2Cl2 and pure CH2Br2 are 133 and 11.4 torr, respectively.

1 answer:

Vinil7 [7]3 years ago

3 0

Answer:

CH2Cl2 mole fraction = 0.833

CH2Br2 mole fraction = 0.167

Explanation:

Step 1: Data given

Number of moles CH2Cl2 = 0.0300 moles

Number of moles CH2Br2 = 0.0500 moles

Temperature = 25 °C

Vapor pressure of pure CH2Cl2 = 133 torr

Vapor pressure of pure CH2Br2 = 11.4 torr

Step 2: Calculate mol fraction in solution

Mol fraction CH2Cl2 = 0.0300 moles / (0.0300 + 0.0700)

Mol fraction CH2Cl2 = 0.300

Mol fraction CH2Br2 = 0.0700 moles / (0.0300 + 0.0700)

Mol fraction CH2Br2 = 0.700

Step 3: Calculate partial pressure

Partial pressure CH2Cl2 = 133 torr * 0.300 = 39.9 torr

Partial pressure CH2Br = 11.4 torr * 0.700 =7.98 torr

Step 4: Calculate the total pressure

total pressure = 39.9 + 7.98 = 47.88 torr

Step 5: Calculate mol fraction

partial pressure / total pressure = mole fraction

CH2Cl2 mole fraction = 39.9 / 47.88 = 0.833

CH2Br2 mole fraction = 7.98 / 47.88 = 0.167

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