using the following table, calculate the freezing point and boiling point of eac

Question

using the following table, calculate the freezing point and boiling point of each of the following solutions

Water H2O = bp 100.0 degress C / Kb 0.51 degrees (C/m) / 0.0 degrees C / Kf 1.86 degrees (C/m)

Ethanol, C2H5OH = bp 78.4 degrees C / Kb 1.22 degrees (C/m) / -114.6 degrees C / Kf 1.99 degrees (C/m)

Chloroform CHCl3 = bp 61.2 degrees C / Kb 3.63 degress (C/m) / -63.5 degress C / Kf 4.68 degrees (C/m)


a. 0.30m glucose in ethanol

b. 20.0g of decane, C10H22, in 45.5 of chloroform CHCl3

c. 0.45 mol ethanol glycol and .15 mol KBr in 150g water

Please help me, and show how you got the answers.

Explanation / Answer

Use the formula delta T = imK where i is the van't Hoff factor, m is molality (moles/Kg) and K is the constant for that liquid. For the first one (a) 0.41 molal glucose in ethanol freezing point. So, the solvent is ethanol and that means the normal freezing point is -114.6, the Kf is 1.99, and for glucose i = 1 because it does not dissociate. The molality is given as 0.41. Plugging numbers into the equation we get... delta T = imk = (1)(0.41)(1.99) = 0.816 degrees. Since the normal freezing point is -114.6 and the change in freezing point is 0.816 degrees, the new freezing point will be -115.4 degrees. Do the others the same way. For (b) decane the i will = 1 and for (c) ethylene glycol the i = 1 and for KBr the i = 2, so total i = 3 in this case

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