An antifreeze solution is prepared from \(222.6 \mathrm{~g}\) of ethylene glycol \(\left[\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{OH})_{2}\right]\) and \(200 \mathrm{~g}\) of water. Calculate the molality of the solution. If the density of the solution is \(1.072 \mathrm{~g} \mathrm{~mL}^{-1}\) then what shall be the molarity of the solution?

Short Answer

Expert verified
The molality of the ethylene glycol solution is approximately \(3.583 \, \text{molal}\). The molarity of the solution is approximately \(3.345 \, \text{M}\).

Step by step solution

01

Calculate the number of moles of ethylene glycol

Use the molar mass of ethylene glycol \(62.07 \frac{g}{mol}\) to find the number of moles by dividing the mass of the substance by its molar mass: \[\text{moles of ethylene glycol} = \frac{222.6 \, g}{62.07 \, \frac{g}{mol}}.\]
02

Calculate the molality of the solution

Molality \(m\) is defined as the number of moles of solute per kilogram of solvent. Therefore, calculate molality using the formula: \[m = \frac{\text{moles of ethylene glycol}}{\text{mass of water in kg}}.\] Note that the mass of water needs to be converted from grams to kilograms.
03

Calculate the molarity of the solution

Molarity \(M\) is the number of moles of solute per liter of solution. To find the molarity, use the total mass of the solution to calculate its volume with the given density. The volume \(V\) in liters is found by dividing the mass of the solution by its density (converted to \(\frac{g}{L}\)). Calculate the total mass of the solution by adding the mass of ethylene glycol and water. The molarity is found using the formula: \[M = \frac{\text{moles of ethylene glycol}}{V}.\]

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Key Concepts

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

Molality
Molality, often denoted as 'm', is a measure of the concentration of a solute in a solution. Specifically, it refers to the number of moles of solute per kilogram of solvent. Unlike molarity, molality is not affected by temperature changes because mass doesn't change with temperature.

To calculate molality, you first need to find the number of moles of the solute, which can be done by dividing its mass by its molar mass. Then, you divide the moles of solute by the mass of the solvent in kilograms. For instance, in the antifreeze example provided, you would divide the mass of ethylene glycol by its molar mass to find its moles and then divide by the mass of water in kilograms.
Molarity
Molarity, represented by 'M', is another way to describe the concentration of a solution. It is defined as the number of moles of solute divided by the volume of the solution in liters. Unlike molality, molarity is temperature-dependent because volume can change with temperature.

When calculating molarity, as in the antifreeze example, you'll first ascertain the total number of moles of the solute. Then you calculate the volume of the solution using its total mass and the given density. Finally, you'll divide the number of moles of solute by this volume in liters. This gives you the molarity of the solution, allowing you to understand how concentrated the solute is within the total volume of the solution.
Solutions Concentration
The concentration of a solution is a measure of how much solute is dissolved in a given quantity of solvent or solution. There are various ways to express this, with molality and molarity being two of them.

The concentration can affect properties like boiling point, freezing point, and vapor pressure. It's important in various applications, from industrial processes to biological systems. When preparing solutions in a lab or industry, precise calculations of concentration are essential to ensure reactions proceed correctly or products meet quality standards.
Chemical Calculations
Chemical calculations are used to quantify substances and their changes during chemical reactions. When dealing with solutions, we often need to calculate quantities like the number of moles, molarity, and molality to name a few.

These calculations require a solid understanding of concepts like molar mass, Avogadro's number, and the relationships between units of mass, volume, and moles. They are pivotal for preparing solutions of precise concentrations, predicting the outcome of reactions, and scaling up processes from the lab to industrial production.

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Most popular questions from this chapter

Calculate the moles of \(\mathrm{H}_{2} \mathrm{O}\) vapours formed if \(1.57\) mole of \(\mathrm{O}_{2}\) are used in presence of excess of \(\mathrm{H}_{2}\) for the given change, $$ 2 \mathrm{H}_{2}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O} $$

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