Question

How many grams of ethylene glycol $\left({\mathrm{C}}_2{\mathrm{H}}_6{\mathrm{O}}_2\right)$ must be added to $1.00\mathrm{kg}$ of water to produce a solution that freezes at $-{5.00}^{\circ }\mathrm{C}?$

Short answer

To produce a solution that freezes at $-{5.00}^{\circ }C$, approximately $167000g$ of ethylene glycol must be added to $1.00kg$ of water.

Step-by-step solution

Finding the cryoscopic constant of water

The cryoscopic constant of water, $K_f$ is given and it's equal to $1.86\frac{{}^{\circ }C}{mol/kg}$.

Calculate the freezing point depression

We have the freezing point of the solution $-{5.00}^{\circ }C$ and the freezing point of pure water is $0^{\circ }C$. The freezing point depression, ΔT, is the difference between these two values: $$\Delta T=T_f-T_{f_{water}}=(-5.00)-0=-5.00{}^{\circ }C$$

Calculate the molality of the solution

Now we can use the freezing point depression equation to find the molality (moles of solute per kilogram of solvent) of the solution: $$molality=\frac{\Delta T}{K_f}=\frac{-5.00}{1.86}=-2.69mol/kg$$ This is a negative value, but molality must be positive, so we will take the absolute value: $$molality=2.69mol/kg$$

Find the mass of ethylene glycol

We now have the molality of the solution, $2.69mol/kg$. To find the mass of ethylene glycol to be added, we need to consider the molecular weight of ethylene glycol, which is around $62.07g/mol$. We have the solvent (water) mass of $1.00kg=1000g$. So the amount of ethylene glycol needed can be calculated as: $$mas{s}_{ethyleneglycol}=molality\times molecularweight\times solventmass=2.69\times 62.07\times 1000$$ $$mas{s}_{ethyleneglycol}=167000g$$ Thus, to produce a solution that freezes at $-{5.00}^{\circ }C$, approximately $167000g$ of ethylene glycol must be added to $1.00kg$ of water.

Key concepts

Cryoscopic Constant

Understanding the cryoscopic constant is essential for studying freezing point depression in solutions. This constant, often represented as K_f, is a unique property for each solvent, indicating how much the freezing point will decrease for every mole of solute particles added per kilogram of the solvent. For water, this value is typically 1.86°C•kg/mol, which means for every one mole of the solute dissolved in one kilogram of water, the freezing point drops by 1.86 degrees Celsius.

The utility of K_f becomes apparent in calculations when we want to predict or analyze the freezing point of a given solution. By multiplying the cryoscopic constant by the molality of the solution—a measure of the concentration of the solute—we can find the depression in the freezing point. The ability to predict freezing points is crucial in many applications, from formulating antifreeze mixtures to crafting ice cream recipes with specific textural qualities.

Molality

When working with solutions and considering how a solute affects the properties of a solvent, molality is a vital concentration unit to comprehend. Defined as the number of moles of solute per kilogram of solvent, molality is written in units of mol/kg. What distinguishes molality from molarity is that molality involves the mass of the solvent rather than the volume of the whole solution, making it particularly useful in situations involving temperature changes—since mass does not change with temperature, whereas volume can.

To calculate molality, one would divide the number of moles of solute by the mass of the solvent in kilograms. Consistent with the properties of a colligative property, molality does not depend on the identity of the solute, but rather on the quantity. This characteristic is particularly relevant when determining the impact of a solute on the freezing point of a solvent, as seen in our textbook exercise.

Colligative Properties

Freezing point depression is one example of colligative properties, which are physical properties of solutions that depend on the ratio of the number of solute particles to the number of solvent molecules in the solution, and not on the nature of the chemical species present. These properties include vapor pressure lowering, boiling point elevation, and osmotic pressure.

Colligative properties arise because the presence of solute molecules interferes with the interactions between solvent molecules. For instance, with freezing point depression, the solute's presence disrupts the formation of the ordered structure of ice, thereby lowering the temperature at which the solution can solidify. This effect can be quantified and predicted using the molality of the solution and the solvent's respective cryoscopic constant. It's important to note that for solutions with non-volatile solutes, like many salts or sugars, the impact on the freezing point is proportional to the number of particles dissolved, irrespective of their specific chemical properties. This feature makes colligative properties immensely useful in a wide variety of scientific and industrial calculations.