Q: Which equation correctly represents the calculation for freezing point depression?

$$\Delta T_f = i \times K_f \times m$$

Q: A 0.15 m NaCl solution and a 0.10 m HCl solution are compared. Which has the lower freezing point and why?

0.15 m NaCl, because its effective molality ($$m \times i) is $$higher.

Question 1

Which of the following statements best explains why the freezing point of a 1.00 molal solution of Na₂SO₄ is lower than that of a 1.00 molal solution of sucrose?

Accepted Answer

Na₂SO₄ dissociates into more ions, increasing the number of solute particles and thus causing a greater freezing point depression.

Question 2

A 0.222 m aqueous solution of sucrose (C₁₂H₂₂O₁₁) has a boiling point elevation constant $$K_b = 0.512\ \mathrm{^\circ C/m}. $$What is the boiling point of the solution?

Accepted Answer

$$100.114\ \mathrm{^\circ C}$$

Question 3

Which of the following best describes the van 't Hoff factor ($$i) $$for a 0.122 m aqueous solution of KI?

Accepted Answer

$$i = 2$$, because KI dissociates into K$$^+$$ and I$$^-$$ ions in water.

Question 4

A 25.0 mass percent solution of ethylene glycol ($$\mathrm{C_2H_6O_2}) in $$water is prepared. If $$K_b = 0.512\ \mathrm{^\circ C/m}$$ and $$K_f = 1.86\ \mathrm{^\circ C/m}$$, what is the freezing point of the solution?

Accepted Answer

$$-9.99\ \mathrm{^\circ C}$$

Question 5

A solution contains 0.600 kg of chloroform ($$\mathrm{$$CHCl_3$$}) $$and 42.0 g of eucalyptol ($$\mathrm{C$$_{10}$$H_{18}O$$}$$). Given K_f$ for $$\mathrm{$$CHCl_3$$}$$ is 4.68$$\ \mathrm{^\circ C/m}$$ and its normal freezing point is -63.5$$\ \mathrm{^\circ C}$, what is the freezing point of the solution?

Accepted Answer

$$-65.6\ \mathrm{^\circ C}$$

Question 6

Arrange the following aqueous solutions in order of their expected freezing points, from lowest to highest: 0.050 m CaCl₂, 0.15 m NaCl, 0.10 m HCl, 0.050 m HC₂H₃O₂, 0.10 m C₁₂H₂₂O₁₁.

Accepted Answer

NaCl \u003c CaCl₂ \u003c HCl \u003c C₁₂H₂₂O₁₁ \u003c HC₂H₃O₂

Question 7

A solution is made by dissolving 0.250 g of a nonvolatile, non-electrolyte solute in 40.0 g of CCl₄. The boiling point of the solution increases by $$0.357\ \mathrm{^\circ C}. If$$ $$K_b = 5.02\ \mathrm{^\circ C/m}$$, what is the molar mass of the solute?

Accepted Answer

$$87.9\ \mathrm{g/mol}$$

Question 8

When 0.186 g of an organic substance is dissolved in 22.01 g of camphor ($$K_f = 40.0\ \mathrm{^\circ C/m}$$, $$T_{	ext{melt}} = 179.8\ \mathrm{^\circ C}$$), the freezing point drops to $$176.7\ \mathrm{^\circ C}. $$What is the approximate molar mass of the solute?

Accepted Answer

$$110\ \mathrm{g/mol}$$

Question 9

A compound called pyrene has the empirical formula $$\mathrm{C_8H_5}. $$When 4.04 g of pyrene is dissolved in 10.00 g of benzene, the boiling point of the solution is $$85.1\ \mathrm{^\circ C}$$ ($$K_b = 2.53\ \mathrm{^\circ C/m}$$, normal BP $$= 80.1\ \mathrm{^\circ C}). $$What is the molecular formula of pyrene?

Accepted Answer

$$\mathrm{C$$_{16}$$H_{10}}$$

Question 10

A mass of CaCl₂, when dissolved in 100.00 g of water, gives a freezing point of $$-5.0\ \mathrm{^\circ C}. $$What mass of glucose ($$\mathrm{C_6H_{12}O_6$}) $$would give the same freezing point depression? ($$K_f = 1.86\ \mathrm{^\circ C/m}$$, $$M_{	ext{glucose}} = 180.18\ \mathrm{g/mol})$$

Accepted Answer

$$48.4\ \mathrm{g}$$

Question 11

Which of the following best explains why CaCl₂ is a more effective antifreeze than glucose when added to water?

Accepted Answer

CaCl₂ dissociates into three ions, increasing the number of solute particles and thus causing a greater freezing point depression.

Question 12

A student dissolves 0.050 mol of acetic acid ($$\mathrm{HC_2H_3O_2}) in 1.00 kg of $$water. Compared to a 0.050 molal solution of NaCl, which will have a higher freezing point and why?

Accepted Answer

Acetic acid, because it is a weak acid and does not fully dissociate, resulting in fewer particles.

Question 13

If a nonvolatile, non-electrolyte solute is added to a solvent, which of the following changes will occur?

Accepted Answer

The boiling point will increase and the freezing point will decrease.

Question 14

A 0.10 m solution of HCl and a 0.10 m solution of C₁₂H₂₂O₁₁ are prepared. Which solution will have a lower freezing point and why?

Accepted Answer

HCl, because it is a strong acid and dissociates into two ions, increasing the effective molality.

Question 15

Which equation correctly represents the calculation for freezing point depression?

Accepted Answer

$$\Delta T_f = i 	imes K_f 	imes m$$

Question 16

A solution is prepared by dissolving 0.2723 mol of a non-electrolyte in 0.600 kg of CHCl₃. If $$K_f = 4.68\ \mathrm{^\circ C/m}$$, what is the expected freezing point depression?

Accepted Answer

$$2.12\ \mathrm{^\circ C}$$

Question 17

A 0.15 m NaCl solution and a 0.10 m HCl solution are compared. Which has the lower freezing point and why?

Accepted Answer

0.15 m NaCl, because its effective molality ($$m 	imes i) is $$higher.

Question 18

Which of the following is a correct step in determining the molar mass of an unknown solute using boiling point elevation data?

Accepted Answer

Calculate the molality from the observed $$\Delta T_b$$ using $$m = \frac{\Delta T_b}{i K_b}.$$

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