Chemisty

A solution is prepared by dissolving 3.44 g of sucrose, C12H22O11, in 118 g of water.

a) What is the molality of the resulting solution?

b) What is the mole fraction of water in this solution?

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asked by Kate
  1. molality=molessugar/kgsolvent
    = 3.44/(molmassSucrose*.118)

    Mole fraction moles solute/totalmoles
    where moles solute=3.44/molmassSucrose
    moles water=18/118

    and total moles=moles solute+moles water.

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  2. Just a slight correction to a typo above.
    mols H2O = 118/18

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  3. B)
    moles of h2o=118g/18(g/mol)=6.5556 mol
    moles of c12h22O11= 3.44g/342.3(g/mol)=0.10049 mol
    6.5556 mol/(6.5556 mol + 0.10049 mol= 0.984902 mol

    A)
    M= (0.10049)/(0.00344 kg + 0.118 kg)= 0.82749 m c12h22O11

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    posted by Kate

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