Question

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 

Answer 1

mole fraction x(FeCl2) = n(FeCl2)/(n(H2O) + n(FeCl2)) = 2.736/(2.736+1000/18.02) = 0.0470