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

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  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 = XCH2Cl2*PoCH2Cl2 and
    pp CCl4 = XCCl4*PoCCl4
    Where Po is the normal vapor pressure of either CH2Cl2 or CCl4.

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