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Chapter 11: Problem 67

What mass of glycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\right)\), a nonelectrolyte, must be dissolved in \(200.0 \mathrm{~g}\) water to give a solution with a freezing point of \(-1.50^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
14.84 g of glycerin must be dissolved in 200.0 g of water to give a solution with a freezing point of -1.50 掳C.

Step by step solution

01

Calculate the change in freezing point

First, we need to calculate the change in freezing point (螖Tf). The freezing point of pure water is 0 掳C, and we are given that the freezing point of the solution is -1.50 掳C. Therefore, the change in freezing point can be found by: \(螖T_f = \) Initial freezing point - Final freezing point \(螖T_f = 0 - (-1.50)\) \(螖T_f = 1.50 \mathrm{~掳C}\)
02

Calculate the molality

Next, we use the freezing point depression equation while taking i = 1 (as glycerin is a nonelectrolyte): \(螖T_f = K_f * \textrm{molality}\) \(1.50 = 1.86 * \textrm{molality}\) Now, we can solve for molality: \(\textrm{molality} = \frac{1.50}{1.86}\) \(\textrm{molality} = 0.806 \mathrm{~mol/kg}\)
03

Use molality to find the moles of glycerin

Now that we have found the molality, we can use it to find the moles of glycerin. Molality is defined as the moles of solute (glycerin) per kg of solvent (water). We know that, there are 200.0 g of water, which can be converted to kg: \(200.0 \mathrm{~g} = 0.200 \mathrm{~kg}\) Now, we can calculate the moles of glycerin: \(0.806 \mathrm{~mol/kg} * 0.200 \mathrm{~kg} = 0.1612 \mathrm{~mol}\)
04

Calculate the mass of glycerin

Finally, we can find the mass of glycerin by multiplying the moles of glycerin with the molar mass of glycerin. The molar mass of glycerin (C3H8O3) can be found by adding the molar masses of its elements: 3C + 8H + 3O: \(3(12.01) + 8(1.008) + 3(16.00) = 92.09 \mathrm{~g/mol}\) Now, we multiply the moles of glycerin by the molar mass: \(0.1612 \mathrm{~mol} * 92.09 \mathrm{~g/mol} = 14.84 \mathrm{~g}\) So, 14.84 g of glycerin must be dissolved in 200.0 g of water to give a solution with a freezing point of -1.50 掳C.

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Key Concepts

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

Molality
Molality is a measure of the concentration of a solute in a solution, which is crucial for understanding how substances dissolve. Defined as the number of moles of solute per kilogram of solvent, molality is different from molarity, which is moles per liter of solution. This distinction is important because molality does not change with temperature, as it depends on mass, not volume.
To calculate molality, you need the mass of the solvent and the amount of solute.
  • First, convert the mass of the solvent from grams to kilograms.
  • Then, divide the moles of solute by the kilograms of solvent to get molality.
In the exercise, the solution had a molality of 0.806 mol/kg, indicating 0.806 moles of glycerin dissolved in each kilogram of water.
Nonelectrolyte Solutions
Nonelectrolyte solutions are those in which the solutes do not dissociate into ions when dissolved in a solvent. These solutions are different from electrolytes, which do dissociate. Substances like glycerin, sugar, and most organic compounds are common nonelectrolytes.
Because they do not produce ions, nonelectrolyte solutions do not conduct electricity. This is a key characteristic, especially important in chemical and physical processes.
  • When determining colligative properties like freezing point depression, nonelectrolytes are considered to have a van 't Hoff factor (i) of 1.
  • In the original exercise, glycerin was used as a nonelectrolyte, meaning the value of i in the freezing point depression equation is 1.
Colligative Properties
Colligative properties are properties of solutions that depend on the number of solute particles in the solution, not the type. They include:
  • Boiling point elevation
  • Freezing point depression
  • Vapor pressure lowering
  • Osmotic pressure
Freezing point depression, a specific colligative property, occurs because solute particles disrupt the formation of the solid structure of the solvent, requiring lower temperatures to freeze. The extent of the freezing point depression is directly proportional to the molality of the solution.
In the exercise, the freezing point was decreased by 1.5 掳C due to the presence of dissolved glycerin, illustrating the effect of colligative properties.

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Most popular questions from this chapter

Patients undergoing an upper gastrointestinal tract laboratory test are typically given an X-ray contrast agent that aids with the radiologic imaging of the anatomy. One such contrast agent is sodium diatrizoate, a nonvolatile water-soluble compound. A \(0.378 \mathrm{~m}\) solution is prepared by dissolving \(38.4 \mathrm{~g}\) sodium diatrizoate (NaDTZ) in \(1.60 \times 10^{2} \mathrm{~mL}\) water at \(31.2^{\circ} \mathrm{C}\) (the density of water at \(31.2^{\circ} \mathrm{C}\) is \(\left.0.995 \mathrm{~g} / \mathrm{mL}\right)\). What is the molar mass of sodium diatrizoate? What is the vapor pressure of this solution if the vapor pressure of pure water at \(31.2^{\circ} \mathrm{C}\) is \(34.1\) torr?

Anthraquinone contains only carbon, hydrogen, and oxygen. When \(4.80 \mathrm{mg}\) anthraquinone is burned, \(14.2 \mathrm{mg} \mathrm{CO}_{2}\) and \(1.65 \mathrm{mg} \mathrm{H}_{2} \mathrm{O}\) are produced. The freezing point of camphor is lowered by \(22.3^{\circ} \mathrm{C}\) when \(1.32 \mathrm{~g}\) anthraquinone is dissolved in \(11.4 \mathrm{~g}\) camphor. Determine the empirical and molecular formulas of anthraquinone.

For an acid or a base, when is the normality of a solution equal to the molarity of the solution and when are the two concentration units different?

How would you prepare \(1.0 \mathrm{~L}\) of an aqueous solution of sodium chloride having an osmotic pressure of \(15 \mathrm{~atm}\) at \(22^{\circ} \mathrm{C} ?\) Assume sodium chloride exists as \(\mathrm{Na}^{+}\) and \(\mathrm{Cl}^{-}\) ions in solution.

A solution of sodium chloride in water has a vapor pressure of \(19.6\) torr at \(25^{\circ} \mathrm{C}\). What is the mole fraction of solute particles in this solution? What would be the vapor pressure of this solution at \(45^{\circ} \mathrm{C} ?\) The vapor pressure of pure water is \(23.8\) torr at \(25^{\circ} \mathrm{C}\) and \(71.9\) torr at \(45^{\circ} \mathrm{C}\) and assume sodium chloride exists as \(\mathrm{Na}^{+}\) and \(\mathrm{Cl}^{-}\) ions in solution.
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