Molality of iron(II) chloride in a solution is 2.736 mol·kg-1.
What is the mole fraction of FeCl2 in the solution?
M(FeCl2) =126.75 g·mol-1, M(H2O) =18.02 g·mol-1
Present your numerical answer to 3 significant figures
mole fraction x(FeCl2) = n(FeCl2)/(n(H2O) + n(FeCl2)) = 2.736/(2.736+1000/18.02) = 0.0470
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