part B
dT = Kf x m, where d = delta (change), Kf is the freezing point
depression constant, and m = molality.
[0-(-2.35)]oC = 1.86 oC/m x m
m = 2.35 oC / 1.86 oC/m = 1.263 molal
part C
dT = Kb x m, where d = delta (change), Kb is the boiling point
elevation constant, and m = molality.
[102.56-100.00] oC = 0.512 oC/m x m
m = 2.56 oC / 0.512 C/m = 5.00 molal
part b and c please ( 5 of 15 A Review | Constants Periodic Table Part...
A solution of water (Kf=1.86 ∘C/m) and glucose freezes at − 2.35 ∘C. What is the molal concentration of glucose in this solution? Assume that the freezing point of pure water is 0.00 ∘C. Express your answer to three significant figures and include the appropriate units. View Available Hint(s) m m m = nothing nothing Submit Boiling points and molality Similar to the freezing-point depression, the boiling-point elevation ΔTb of a solution is quantitatively related to the molality m and...
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