Four liquids are described in the table below. Use the second column of the tabl

Question

Four liquids are described in the table below. Use the second column of the table to explain the order of their freezing points, and the third column to explain the order of their boiling points For example, select '1' in the second column next to the liquid with the lowest higher freezing point, a point, and so on. so on. In the third column, select 1 next to the liquid with the lowest boling point, "2' next to the liquid with the next higher bolling Note: the density of water is 1.00 g/mL solution 3.2 g of potassium iodide (K?) dssorved in 200, mL of water 3.2 g of sucrose (CizMzz011) ssolved in 200. mL of water 3.2 g of ethylene glycol (Ce0z) dissoived in 200. ml of water 200, mL of pure water freezing point boiling point (choose one)(choose one) v (choose one)(choose one) choose one)Kchoose one) Choose one) (choose one) v

Explanation / Answer

To calculate the freezing point:

The molality of the solution is determined: Case KI: moles KI = 3.2gr * (1 mol / 166gr) = 0.0193 moles Kg H2O = 200 ml * (1 gr / ml) * (1 kg / 1000 gr) = 0.2 Kg m = 0.0193 moles / 0.2 Kg = 0.0964 m The freezing temperature change is calculated: with the freezing point constant Kc = - 1.86 ºC * Kg / mol ?Tc = Kc * m = -1.86ºC * Kg / mol * 0.0964 mol / Kg = -0.1793 ºC and finally we calculate the freezing point of the solution: Tc = 0ºC + (-0.1793ºC) = -0.1793ºC In the same way it is done with each component and the order of freezing point from lowest to highest is: KI: 2 Sucrose: 3 Ethylene glycol: 1 Pure Water: 4 To calculate the boiling point: We already have the molality data for the KI, then we directly calculate the change of the boiling point: with Ke = 0.52 ° C * Kg / mol ? Te = Ke * m = 0.52 ºC * Kg / mol * 0.0964 mol / Kg = 0.0501 ºC Te = 100ºC + 0.0501ºC = 100.0501ºC The rest of the solutes are calculated in the same way and we have the order from least to greatest: KI: 3 Sucrose: 2 Ethylene glycol: 4 Pure Water: 1

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