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What would be the freezing and boiling points of a solution prepared by
dissolving 1.00g of benzoic acid (C6H5CO2H) in 10.0g of benzene?

  • Chemistry - ,

    mols acid = grams/molar mass
    m acid = mols/kg solvent
    delta T = Kf*m
    Solve for T and add to normal freezing point of benzene to find the new f.p.

    Do the same for b.p.
    mols acid is same
    m acid is same
    delta T = Kb*m
    Add delta T to normal b.p. to find new b.p.

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