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Ch.13 - Solutions & Their Properties
Chapter 13, Problem 14

An aqueous solution of a certain organic compound has a density of 1.063 g/mL, an osmotic pressure of 12.16 atm at 25.0 °C, and a freezing point of -1.03 °C. The compound is known not to dissociate in water. What is the molar mass of the compound?

Verified step by step guidance
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Step 1: Use the osmotic pressure formula \( \Pi = iMRT \) to find the molarity (M) of the solution. Since the compound does not dissociate, the van't Hoff factor \( i = 1 \). Rearrange the formula to solve for M: \( M = \frac{\Pi}{RT} \), where \( R = 0.0821 \text{ L atm/mol K} \) and \( T = 298.15 \text{ K} \) (25.0 °C in Kelvin).
Step 2: Calculate the molality (m) of the solution using the freezing point depression formula \( \Delta T_f = iK_f m \). Since the compound does not dissociate, \( i = 1 \). Rearrange the formula to solve for m: \( m = \frac{\Delta T_f}{K_f} \), where \( \Delta T_f = 1.03 \text{ °C} \) and \( K_f \) is the cryoscopic constant for water (1.86 °C kg/mol).
Step 3: Use the density of the solution to find the mass of the solvent. Assume 1 L of solution for simplicity, which has a mass of 1063 g (since density = 1.063 g/mL). Subtract the mass of the solute (found from molality) to find the mass of the solvent.
Step 4: Convert the molality to moles of solute using the mass of the solvent in kg. Use the relationship \( m = \frac{\text{moles of solute}}{\text{kg of solvent}} \) to find the moles of solute.
Step 5: Calculate the molar mass of the compound by dividing the mass of the solute (from the density and volume of solution) by the moles of solute (from the molality calculation). Use the formula \( \text{Molar mass} = \frac{\text{mass of solute}}{\text{moles of solute}} \).

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Osmotic Pressure

Osmotic pressure is the pressure required to prevent the flow of solvent into a solution through a semipermeable membrane. It is directly proportional to the molarity of the solute and can be calculated using the formula π = iCRT, where π is the osmotic pressure, i is the van 't Hoff factor (which is 1 for non-dissociating compounds), C is the molarity, R is the ideal gas constant, and T is the temperature in Kelvin.
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Osmotic Pressure Formula

Freezing Point Depression

Freezing point depression is a colligative property that describes the lowering of the freezing point of a solvent when a solute is added. The extent of freezing point depression can be calculated using the formula ΔTf = iKf m, where ΔTf is the change in freezing point, Kf is the freezing point depression constant of the solvent, and m is the molality of the solution. This concept is essential for determining the molar mass of the solute.
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Freezing Point Depression

Molar Mass Calculation

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). To find the molar mass of a solute from colligative properties, one can rearrange the formulas for osmotic pressure or freezing point depression to solve for molar mass, using the measured values of osmotic pressure, freezing point change, and the properties of the solvent. This calculation is crucial for identifying the molecular weight of the organic compound in the question.
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Molar Mass Calculation Example
Related Practice
Textbook Question
12. A solution contains 4.08 g of chloroform 1CHCl32 and 9.29 g of acetone 1CH3COCH32. The vapor pressures at 35 °C of pure chloroform and pure acetone are 295 torr and 332 torr, respectively. Assuming ideal behavior, calculate the vapor pressure above the solution. (LO 13.12) (a) 256 torr (b) 314 torr (c) 325 torr (d) 462 torr
Textbook Question
How many kilograms of ethylene glycol (automobile anti- freeze, C2H6O2) dissolved in 3.55 kg of water are needed to lower the freezing point of water in an automobile radiator to - 22.0 °C? The molal freezing point depression constant for water is Kf = 1.86 1°C kg2>mol. (LO 13.13) (a) 0.865 kg (b) 0.0420 kg (c) 9.01 kg (d) 2.61 kg
Textbook Question
Hemoglobin is a large molecule that carries oxygen in the body. An aqueous solution that contains 2.61 g of hemoglo- bin in 100.0 mL has an osmotic pressure of 7.52 mmHg at 25 °C. What is the molar mass of the hemoglobin? Assume hemoglobin does not dissociate in water. (LO 13.15) (a) 1.96 * 103 g>mol (b) 84.8 g/mol (c) 6.45 * 104 g>mol (d) 3.65 * 103 g>mol
Textbook Question
The following phase diagram shows part of the vapor- pressure curves for a pure liquid (green curve) and a solution of the first liquid with a second volatile liquid (red curve).

(a) Is the boiling point of the second liquid higher or lower than that of the first liquid?
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Textbook Question
The following phase diagram shows part of the liquid–vapor phase-transition boundaries for pure ether and a solution of a nonvolatile solute in ether.

(b) What is the approximate mo#lal concentration of the sol-ute? [Kb for ether is 2.021°C kg2>mol.]