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 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 freezing pointboiling point (choose one) (choose one)choose one) solution choose one) 8.1 g of sodium bromide (NaBr) dissolved in 450. mL of water 8.1 g of potassium nitrate (KNO2) dissolved in 450. mL of water 8.1 g of glucose (CsH120s) dissolved in 450. mL of water 450. mL of pure water choose one) (choose one): (choose one) (choose one):

Explanation / Answer

1.

NaBr (aq.) ----------> Na+ (aq.) + Br- (aq.)

So, van't Hoff factor , i = 2

(a) depreesion in freezing point = i * 1000 * Kf * w / ( W * m )

0.0 - Tf = 2 * 1000 * 1.86 * 8.1 / (450. * 103)

Tf = freezing point = - 0.650 0C

(b)

Elevation in boiling point = i * 1000 * Kb * w / (W*m)

Tb - 100 = 2 * 1000 * 0.512 * 8.1 / (450. * 103)

Tb = boiling point = 100.179 0C

2.

KNO3 (aq.) -----------> K+ (aq.) + NO3- (aq.)

SO, van't factor, i = 2

(a)

0.0 - Tf = 2 * 1000 * 1.86 * 8.1 / (450. * 101)

Tf = - 0.663 0C

(b)

Tb - 100 = 2 * 1000 * 0.512 * 8.1 / (450. * 101)

Tb = 100.182 0C

3.

(a)

C6H12O6 does not undergo dissociation in water.

So, van't Hoff factor = 1

0.0 - Tf = 1 * 1000 * 1.86 * 8.1 / (450. * 180)

Tf = - 0.186 0C

(b)

Tb -100 = 1. * 1000 * 0.512 * 8.1 / (450. * 180)

Tb = 100.0512 0C

4.

Freezing point of pure water = 0.0 0C

Boiling point of pure water = 100 0C

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