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Chapter 9: Problem 55

Equal volumes of ethylene glycol (molar mass = 62 ) and water (molar mass \(=18\) ) are mixed. The depression in freezing point of water is (given \(\mathrm{K}_{\mathrm{f}}\) of water \(=1.86 \mathrm{~K}\) mol-1 \(\mathrm{kg}\) and specific gravity of ethylene glycol is 1.11) (a) \(0.0033\) (b) \(0.033\) (c) \(0.33\) (d) \(33.3\)

Short Answer

Expert verified
The depression in freezing point is 33.3 K, option (d).

Step by step solution

01

Calculate Molality of the Solution

First, determine the mass of ethylene glycol and water in the mixture. Assume 100 mL of each liquid for simplicity. The mass of ethylene glycol is calculated as follows:\[ \text{mass of ethylene glycol} = \text{volume} \times \text{density} = 100 \text{ mL} \times 1.11 \text{ g/mL} = 111 \text{ g} \]The molar mass of ethylene glycol is 62, so the number of moles is:\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{111}{62} \approx 1.79 \text{ moles} \]The mass of water is:\[ \text{mass of water} = \text{volume} \times \text{density} = 100 \text{ mL} \times 1 \text{ g/mL} = 100 \text{ g} \]Convert this mass into kilograms for the molality calculation:\[ \text{mass of water in kg} = 0.1 \text{ kg} \]Now, calculate the molality:\[ \text{molality (m)} = \frac{\text{moles of ethylene glycol}}{\text{mass of water in kg}} = \frac{1.79}{0.1} = 17.9 \text{ mol/kg} \]
02

Calculate Depression in Freezing Point

Using the molality obtained, apply it in the formula for freezing point depression:\[ \Delta T_f = i \times K_f \times m \]where \( \Delta T_f \) is the depression in freezing point, \( i = 1 \) (since ethylene glycol does not dissociate), \( K_f = 1.86 \text{ K mol}^{-1} \text{ kg} \), and \( m = 17.9 \text{ mol/kg} \). Thus,\[ \Delta T_f = 1 \times 1.86 \times 17.9 = 33.3 \text{ K} \]
03

Conclusion: Identify the Correct Answer

The calculated depression in freezing point is 33.3 K, which matches the option (d).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Molality
Molality is a concept used to express the concentration of a solution. It differs from molarity, which is based on the volume of the solution, while molality is dependent on the mass of the solvent.
Molality is tremendously convenient for calculations involving temperature changes because it does not vary with temperature. This is crucial for accurate freezing point depression calculations.

To calculate molality, use the formula:
  • The number of moles of solute, in this case, ethylene glycol.
  • The mass of the solvent, which is water, in kilograms.
In our exercise, we have equal volumes of ethylene glycol and water. By finding the mass and converting to kilograms, we calculate the molality as the number of moles of ethylene glycol divided by the mass of water in kilograms, resulting in 17.9 mol/kg.
Why Ethylene Glycol?
Ethylene glycol is a compound known for its effectiveness in lowering the freezing point of a solution, which makes it a popular antifreeze agent. It's a simple molecule with the formula \(\(\text{C}_2\text{H}_6\text{O}_2\)\), meaning it has two alcohol functional groups.
These functional groups are responsible for hydrogen bonding with water molecules, effectively disrupting the orderly ice structure required for freezing. Thus, less temperature is needed to freeze a water-ethylene glycol mixture compared to pure water.

In the exercise, ethylene glycol is mixed with water, causing a significant depression in the freezing point, as evidenced by the calculated depression of 33.3 K. Its molar mass and specific gravity are crucial for determining the precise number of moles in our calculations.
Applying the Freezing Point Depression Formula
The freezing point depression formula is essential when you need to determine how much the freezing point of a liquid drops after adding a solute. The formula used is:\[ \Delta T_f = i \times K_f \times m \]
  • \( \Delta T_f \) is the change in freezing point.
  • \( i \) is the van 't Hoff factor, representing how many particles the solute dissociates into. For ethylene glycol, \( i = 1 \) because it does not dissociate.
  • \( K_f \) is the cryoscopic constant of the solvent, provided as 1.86 K mol鈦宦� kg in our example for water.
  • \( m \) is the molality of the solution.
This formula combines all these factors to express the lowering of the freezing point resulting from mixing ethylene glycol with water. In our specific case, the conclusion is that the freezing point drops by 33.3 K.

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Most popular questions from this chapter

Which of the following represents the decreasing order of \(\Delta T_{b}\) ? (i) \(0.02 \mathrm{M}\) Glucose (ii) \(0.01 \mathrm{M} \mathrm{KC}_{1}\) (iii) \(0.02 \mathrm{M} \mathrm{CaF}_{2}\) (iv) \(0.01 \mathrm{M} \mathrm{AlCl}_{3}\) a. (iv) \(>(\mathrm{iii})>(\mathrm{ii})=\) (i) b. (iii) \(>(\mathrm{iv})>(\mathrm{ii})=(\mathrm{i})\) c. (iii) \(>(\mathrm{iv})>(\mathrm{ii})>\) (i) d. (iv) \(>(\mathrm{iii})>(\mathrm{ii})>(\mathrm{i})\)

(A): Elevation in boiling point will be high if the molal elevation constant of the liquid is high. (R): Elevation in boiling point is a colligative property.

Which one of the following statements is/are true? a. two sucrose solutions of same molality prepared in different solvents will have the same freezing point depression b. the osmotic pressure ( \(\pi\) ) of a solution is given by the equation \(\pi=\) MRT, where \(M\) is the molarity of the solution c. Raoult's law states that the vapour pressure of a component over a solution is proportional to its mole fraction d. the correct order of osmotic pressure for \(0.01\) M aqueous solution of each compound is \(\mathrm{BaCl}_{2}>\mathrm{KCl}<\mathrm{CH}_{3} \mathrm{COOH}>\) Sucrose

At \(25^{\circ} \mathrm{C}\), the total pressure of an ideal solution obtained by mixing 3 moles of \(\mathrm{A}\) and 2 moles of \(\mathrm{B}\) is 184 torr. What is the vapour pressure (in torr) of pure B at the same temperature? (Vapour pressure of pure \(\mathrm{A}\), at \(25^{\circ} \mathrm{C}\), is 200 torr \()\) (a) 100 (b) 16 (c) 160 (d) 1

Match the following for an ideal solution: Column 1 A. \(\Delta \mathrm{H}\) B. \(\Delta V\) C. A - B interaction D. total vapour pressure Column II (p) observed vapour pressure (q) equals to \(\mathrm{A}-\mathrm{A}\) or \(\mathrm{B}-\mathrm{B}\) (r) zero (s) positive
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