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 = XCCl4*P^oCCl4
Where P^o is the normal vapor pressure of either CH2Cl2 or CCl4.}