The U.S. Food and Drug Administration lists dichloromethane (CH2Cl2) and carbon tetrachloride (CCl4) among the many cancer-causing chlorinated organic compounds. what are the partial pressures of these substances in the vapor above a mixture containing 1.75 mol of CH2Cl2 and 1.95 mol of CCl4 at 23.5 degrees C? the vapor pressures of pure CH2Cl2 and CCl4 at 23.5 degrees C and 352 torr and 118 torr respectively

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To determine the partial pressures of dichloromethane (CH2Cl2) and carbon tetrachloride (CCl4) in the vapor above the mixture, we can use Raoult's law.

According to Raoult's law, the vapor pressure of a component in an ideal solution is directly proportional to its mole fraction in the mixture.

First, we need to calculate the mole fractions (X) of each component:

Mole fraction (X) = Number of moles of the component / Total moles of the mixture

For CH2Cl2:
X(CH2Cl2) = 1.75 mol / (1.75 mol + 1.95 mol) = 0.473

For CCl4:
X(CCl4) = 1.95 mol / (1.75 mol + 1.95 mol) = 0.527

Now, we can calculate the partial pressures (P) of each component using Raoult's law:

P(CH2Cl2) = X(CH2Cl2) * P°(CH2Cl2)
P(CCl4) = X(CCl4) * P°(CCl4)

Given:
P°(CH2Cl2) = 352 torr
P°(CCl4) = 118 torr

P(CH2Cl2) = 0.473 * 352 torr = 166.896 torr
P(CCl4) = 0.527 * 118 torr = 62.086 torr

Therefore, the partial pressures of CH2Cl2 and CCl4 in the vapor above the mixture are approximately 166.896 torr and 62.086 torr, respectively.

To calculate the partial pressures of dichloromethane (CH2Cl2) and carbon tetrachloride (CCl4) in the vapor above the mixture, we will use Dalton's Law of Partial Pressure, which states that the total pressure of a mixture of non-reactive gases is equal to the sum of the pressures each gas would exert if it were present alone.

Let's start by calculating the mole fraction (X) of each substance in the mixture.
Mole fraction can be calculated using the formula:
X = moles of substance / total moles of all substances

Mole fraction of CH2Cl2:
X(CH2Cl2) = moles of CH2Cl2 / (moles of CH2Cl2 + moles of CCl4)
X(CH2Cl2) = 1.75 mol / (1.75 mol + 1.95 mol)
X(CH2Cl2) ≈ 0.473

Mole fraction of CCl4:
X(CCl4) = moles of CCl4 / (moles of CH2Cl2 + moles of CCl4)
X(CCl4) = 1.95 mol / (1.75 mol + 1.95 mol)
X(CCl4) ≈ 0.527

Next, we need to calculate the partial pressure of each substance in the mixture using their respective vapor pressures.

Partial pressure of CH2Cl2:
P(CH2Cl2) = X(CH2Cl2) * vapor pressure of CH2Cl2
P(CH2Cl2) = 0.473 * 352 torr
P(CH2Cl2) ≈ 166.096 torr

Partial pressure of CCl4:
P(CCl4) = X(CCl4) * vapor pressure of CCl4
P(CCl4) = 0.527 * 118 torr
P(CCl4) ≈ 62.186 torr

Therefore, the partial pressure of dichloromethane (CH2Cl2) in the vapor above the mixture is approximately 166.096 torr, and the partial pressure of carbon tetrachloride (CCl4) is approximately 62.186 torr.