What is the molality of an aqueous solution of sucrose that freezes at -3.00 °C? The freezing point lowering constant for water is -1.86 °C/molal.
0.62 molal
1.62 molal
2.37 molal
5.58 molal
de;ta T = Kf*m
To determine the molality of the solution, we can use the formula:
ΔT = -Kf * m
Where:
ΔT = change in freezing point (in this case, ΔT = -3.00 °C)
Kf = freezing point lowering constant (-1.86 °C/molal for water)
m = molality of the solution (what we are trying to find)
Rearranging the equation, we have:
m = -ΔT / Kf
Substituting the given values, we find:
m = -(-3.00 °C) / (-1.86 °C/molal)
m = 3.00 °C / 1.86 °C/molal
m ≈ 1.61 molal
So, the molality of the aqueous solution of sucrose is approximately 1.61 molal.
Therefore, the closest option from the given choices is 1.62 molal.