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 explairn the order of their boling points. For example, select '1' in the second column next to the liquid with the lowest freezing point. Select '2' in the second column next to the liquid with the next higher freezing point, and so on. In the third column, select '1 next to the liquid with the lowest boiling point, '2" next to the liquid with the next higher boiling point, and so on. Note: the density of water is 1.00 g/mL solution freezing pointboiling point (choose one) ( 7.6 g of sucrose (C12H22011) dissolved in 200. mL of water 7.6 g of potassium sulfate (K2S04) dissolved in 200. mL of water(choose one)(choose one) 7.6 g of potassium acetate (KCH3CO2) dissolved in 200. mL of water 200. mL of pure water (choose one) |choose one)! choose one) 1(lowest)

Explanation / Answer

Freezing point depression

Tf - Ts = i x Kf x m

Part a

0 - Ts = 1 x 1.86 x 7.6/342*0.2

- Ts = 0.20

Ts = - 0.20°C

Tf - Ts = i x Kf x m

Part b

0 - Ts = 3 x 1.86 x 7.6/174.2*0.2

- Ts = 1.217

Ts = - 1.217°C

Part c

0 - Ts = 2 x 1.86 x 7.6/98.15*0.2

Ts = - 1.44°C

Part d

H2O freezing point = 0°C

Rank

7.6 g of sucrose = 3

7.6 g of potassium sulfate = 2

7.6 g of potassium acetate = 1 (lowest)

200 mL water = 4 (highest)

Boiling point elevation

Ts - Tb = i x Kb x m

Part a

Ts - 100 = 1 x 0.512 x 7.6/342*0.2

Ts = 100.0568 °C

Part b

Ts - 100 = 3 x 0.512 x 7.6/174.2*0.2

Ts = 100.335 °C

Part C

Ts - 100 = 2 x 0.512 x 7.6/98.15*0.2

Ts = 100.396 °C

Part d

H2O boiling point = 100°C

Rank

7.6 g of sucrose = 2

7.6 g of potassium sulfate = 3

7.6 g of potassium acetate = 4 (highest)

200 mL water = 1 (lowest)

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