Chemistry Check Please! ?Fixed

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?

B)
moles of h2o=118g/18(g/mol)=6.548 mol

moles of c12h22O11= 3.44g/342.3(g/mol)=0.010 mol

--->6.5556 mol/(6.5556 mol + 0.010 mol= 0.998 mol

A)
M= (0.010)/(0.00344 kg + 0.118 kg)= 0.823 m c12h22O11

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asked by Kate
  1. I think you have missed the point for A.

    For B you haven't been consistent with significant figures and you have different numbers (but close) for the same operation.
    mols H2O = 118/18.0 = 6.556 which I would round to 6.56
    mols C12H22O11 = 3.44/342 = 0.010
    total mols = 6.56+0.01 = 6.57
    6.56/6.57 = 0.998


    For a.
    m stands for molality. M is for molarity.
    m = mols solute/kg solvent
    m = 0.01/0.118 = ?

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