Asked by Monica

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
Answered by DrBob222
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<sub>CH2Cl2</sub>*P<sup>o</sup><sub>CH2Cl2 and
pp CCl4 = X<sub>CCl4</sub>*P<sup>o</sup><sub>CCl4</sub>
Where P<sup>o</sup> is the normal vapor pressure of either CH2Cl2 or CCl4.
There are no AI answers yet. The ability to request AI answers is coming soon!