Determine the freezing point of a solution that contains 0.31 mol of sucrose in 175 g of water

1 answer:

8 0

ΔT= T (solvet freezing point)-T (solution) =T(water freezing point)-T(solution)
ΔT =0⁰C - T(solution)

ΔT= $K_{f} *m$

molality =(number mol solute)/kg solvent = 0.31mol/0.175 kg =(0.31/0.175)m
K(f water)=1.86⁰C/m
ΔT=1.86⁰C/m*(0.31/0.175)m=3.29⁰C
T(solution) =0⁰C - 3.29⁰C =3.29⁰C

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