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A solution is made by dissolving 10.20 grams of glucose (C6H12O6) in 355 grams of water. What is the freezing-point depression of the solvent if the freezing point constant is -1.86 °C/m? Show all of the work needed to solve this problem.

Respuesta :

lulu1991
The freezing point constant means that the temperature depression when adding one mole of the solute. The mole number of the solute is 10.20/180=0.0567 mol. So the answer is 1.86*0.0567=0.105 ℃.

10.20 g C6H12O6 x (1 mole C6H12O6 / 180 g C6H12O6) = 0.0567 mole C6H12O6

Then convert grams of water to kilograms of water by dividing by 1000.  

355 g / 1000 = .355 Kg of H2O

molality (m) = (0.0567 mole C6H12O6 / .355 Kg of H2O) = 0.160 m  

Then plug it in to the formula Δtf = Kfm

Δtf = (-1.86 °C/m) x (0.160 m)  

Δtf = -0.298 °C


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