Asample of glucose ( c6h12o6 ) of mass 8.44 grams is dissolved in 2.11 kg water. what is the freezing point of this solution? the freezing point depression constant, kf , for water is 1.86 °c/mol. (round your answer to the nearest thousandth)

- 0.0413°C ≅ - 0.041°C (nearest thousands).

Explanation:

ΔTf = Kf.m,

Where,

ΔTf is the change in the freezing point.

Kf is the freezing point depression constant (Kf = 1.86 °C/m).

m is the molality of the solution.

Molality is the no. of moles of solute per kg of the solution.

∴ molality (m) = no. of moles of solute/kg of solvent = (0.04685 mol)/(2.11 kg) = 0.0222 m.

∴ ΔTf = Kf.m = (1.86 °C/m)(0.0222 m) = 0.0413°C.

∴ The freezing point of the solution = the freezing point of water - ΔTf = 0.0°C - 0.0413°C = - 0.0413°C ≅ - 0.041°C (nearest thousands).

Hi!

Assuming the 9in is the radius, we would multiply it by 2, or add it to itself. We then get 18 in.

Hope this helps!