The U.S. Food and Drug Administration lists dichloromethane (CH2Cl2) and carbon tetrachloride (CCL4) among the many cancer-causing 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 are 352 torr and 118 torr, respectively. I'm a little confused about how to set it up

Question

The U.S. Food and Drug Administration lists dichloromethane (CH2Cl2) and carbon tetrachloride (CCL4) among the many cancer-causing 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 are 352 torr and 118 torr, respectively. I'm a little confused about how to set it up

Answer 1

You have moles of each.

Find total moles = 1.75 + 1.95 = ??
mole fraction CH2Cl2 = moles CH2Cl2/total moles.

mole fraction CCl4 = moles CCl4/total moles.

partial pressure is pp.
pp CH2Cl2 = X_CH2Cl2*P^o_{CH2Cl2 and

pp CCl4 = X_CCl4*P^o_CCl4

Where P^o is the normal vapor pressure of either CH2Cl2 or CCl4.}

Answer 2

To determine the partial pressures of dichloromethane (CH2Cl2) and carbon tetrachloride (CCl4) in the vapor above the mixture, you can use Raoult's law. Raoult's law relates the partial pressure of a component in a mixture to its vapor pressure and mole fraction in the liquid phase.

Here's how you can set it up:

Step 1: Calculate the mole fraction of CH2Cl2 (X_CH2Cl2) and CCl4 (X_CCl4) in the mixture.
Mole fraction (X) is the ratio of moles of a component to the total moles of all components in the mixture. You can calculate it using the following formulas:

X_CH2Cl2 = (moles of CH2Cl2) / (moles of CH2Cl2 + moles of CCl4)
X_CCl4 = (moles of CCl4) / (moles of CH2Cl2 + moles of CCl4)

Given:
moles of CH2Cl2 = 1.75 mol
moles of CCl4 = 1.95 mol

X_CH2Cl2 = 1.75 / (1.75 + 1.95) = 0.4737
X_CCl4 = 1.95 / (1.75 + 1.95) = 0.5263

Step 2: Use Raoult's law to calculate the partial pressures of CH2Cl2 and CCl4.
According to Raoult's law, the partial pressure of a component (P_partial) is equal to the vapor pressure of the pure component (P_vapor) multiplied by its mole fraction (X) in the mixture.

P_CH2Cl2 = P_vapor_CH2Cl2 * X_CH2Cl2
P_CCl4 = P_vapor_CCl4 * X_CCl4

Given:
P_vapor_CH2Cl2 = 352 torr
P_vapor_CCl4 = 118 torr

P_CH2Cl2 = 352 * 0.4737 ≈ 166.3 torr
P_CCl4 = 118 * 0.5263 ≈ 62.1 torr

So, the partial pressure of CH2Cl2 in the vapor above the mixture is approximately 166.3 torr, and the partial pressure of CCl4 is approximately 62.1 torr.

Remember, Raoult's law assumes ideal behavior and that the components do not react with each other.

Answer 3

To determine the partial pressures of dichloromethane (CH2Cl2) and carbon tetrachloride (CCl4) in the vapor above the mixture, you can use Dalton's Law of Partial Pressures. This law states that the total pressure exerted by a mixture of non-reacting gases is the sum of the partial pressures of each gas.

First, let's calculate the mole fractions of CH2Cl2 and CCl4 in the mixture:

Mole fraction of CH2Cl2 (X_CH2Cl2) = Number of moles of CH2Cl2 / Total moles in the mixture
= 1.75 mol / (1.75 mol + 1.95 mol)
= 1.75 mol / 3.70 mol
≈ 0.473

Mole fraction of CCl4 (X_CCl4) = Number of moles of CCl4 / Total moles in the mixture
= 1.95 mol / (1.75 mol + 1.95 mol)
= 1.95 mol / 3.70 mol
≈ 0.527

Next, use the mole fractions to calculate the partial pressures of each substance:

Partial pressure of CH2Cl2 (P_CH2Cl2) = Mole fraction of CH2Cl2 × Vapor pressure of pure CH2Cl2
= 0.473 × 352 torr
≈ 166.496 torr

Partial pressure of CCl4 (P_CCl4) = Mole fraction of CCl4 × Vapor pressure of pure CCl4
= 0.527 × 118 torr
≈ 62.086 torr

Therefore, the partial pressure of dichloromethane in the vapor above the mixture is approximately 166.496 torr, and the partial pressure of carbon tetrachloride is around 62.086 torr.