Question

A solution is made by dissolving 25.5 grams of glucose (C6H12O6) in 398 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.

I'm stock on this question, can anyone please help me?

Thank you(:

Answer 1

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Answer 2

freezing-point depression is found by using the equation:
\[\Delta T_{B} = i*K_{B}*m\] where i is the ion-factor, KB is the freezing-point constant for a given solvent (here it's water) and m is the molality of the solution. Molality is\[m = \frac{mol solute}{kg solvent}\]
Find the molality of the glucose solution first. Then use the molality to find the DT. The ion-factor for a nonelectrolyte like glucose is always 1, so we can often ignore it.