Question

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

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 °C.

Step-by-step solution

Calculate the change in freezing point

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

Calculate the molality

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

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: \(0.806 \mathrm{~mol/kg} * 0.200 \mathrm{~kg} = 0.1612 \mathrm{~mol}\)

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 °C.

Key concepts

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 °C due to the presence of dissolved glycerin, illustrating the effect of colligative properties.