Problem 23 - Chemistry: An Atoms-Focused Approach

Question

Calculate the osmotic pressure of each of the following aqueous solutions at \(20^{\circ} \mathrm{C}\) a. \(2.39 M\) methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) b. \(9.45 \mathrm{mM} \mathrm{MgCl}_{2}\) c. \(40.0 \mathrm{mL}\) of glycerol \(\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\right)\) in \(250.0 \mathrm{mL}\) of aqueous solution (density of glycerol \(=1.265 \mathrm{g} / \mathrm{mL}\) ) d. \(25 \mathrm{g}\) of \(\mathrm{CaCl}_{2}\) in \(350 \mathrm{mL}\) of solution

Step-by-Step Solution

Answer

a. a 2.39 M methanol solution b. a 9.45 mM magnesium chloride solution c. a solution containing 40.0 mL of glycerol with a density of 1.265 g/mL diluted to 250 mL with water d. a solution prepared by dissolving 25 g of calcium chloride in enough water to make 350 mL of solution The osmotic pressures for the given solutions are: a. 48.02 atm b. 6.67 atm c. 52.16 atm d. 45.57 atm

Step 1: Convert temperature to Kelvin

First, let's convert the temperature from Celsius to Kelvin. The formula for conversion is \(T_{K} = T_{^{\circ} \mathrm{C}} + 273.15\). Given the temperature is \(20^{\circ} \mathrm{C}\), we get: $$T_K = 20 + 273.15 = 293.15 \mathrm{K}$$

Step 2: Calculate osmotic pressure for 2.39 M methanol solution

Since methanol is a non-electrolyte, \(i=1\). We can plug in the values into the osmotic pressure formula: $$\Pi = i \times M \times R \times T = 1 \times 2.39 \times 0.0821 \times 293.15 = 48.02 \mathrm{atm}$$

Step 3: Calculate osmotic pressure for 9.45 mM MgCl2 solution

For \(\mathrm{MgCl}_{2}\) , the van't Hoff factor \(i=3\). First, convert the concentration to M by dividing by \(1000\): $$9.45 \mathrm{mM} = 9.45 \times 10^{-3} \mathrm{M}$$ Now, calculate the osmotic pressure: $$\Pi = i \times M \times R \times T = 3 \times 9.45 \times 10^{-3} \times 0.0821 \times 293.15 = 6.67 \mathrm{atm}$$

Step 4: Calculate osmotic pressure of the glycerol solution

First, calculate the moles of glycerol in the solution. We know its volume and density, so we can obtain its mass: $$\text{mass of glycerol} = \text{volume} \times \text{density} = 40.0 \mathrm{mL} \times 1.265 \mathrm{g/mL} = 50.60 \mathrm{g}$$ Now, we can determine its moles using the molar mass of glycerol (\(92.09 \mathrm{g/mol}\)): $$\text{moles of glycerol} = \frac{50.60 \mathrm{g}}{92.09 \mathrm{g/mol}} = 0.549 \mathrm{mol}$$ Next, calculate the molar concentration (\(\text{total volume} = 250.0 \mathrm{mL} = 0.250 \mathrm{L}\)): $$M = \frac{0.549 \mathrm{mol}}{0.250 \mathrm{L}} = 2.196 \mathrm{M}$$ Glycerol is a non-electrolyte, so \(i=1\). Calculate the osmotic pressure: $$\Pi = i \times M \times R \times T = 1 \times 2.196 \times 0.0821 \times 293.15 = 52.16 \mathrm{atm}$$

Step 5: Calculate osmotic pressure of the CaCl2 solution

First, calculate the moles of \(\mathrm{CaCl}_{2}\) using its molar mass (\(110.98 \mathrm{g/mol}\)): $$\text{moles of CaCl}_2 = \frac{25 \mathrm{g}}{110.98 \mathrm{g/mol}} = 0.225 \mathrm{mol}$$ Next, calculate the molar concentration (\(\text{total volume} = 350 \mathrm{mL} = 0.350 \mathrm{L}\)): $$M = \frac{0.225 \mathrm{mol}}{0.350 \mathrm{L}} = 0.6429 \mathrm{M}$$ \(\mathrm{CaCl}_{2}\) dissociates into \(3\) ions, so \(i=3\). Calculate the osmotic pressure: $$\Pi = i \times M \times R \times T = 3 \times 0.6429 \times 0.0821 \times 293.15 = 45.57 \mathrm{atm}$$ The osmotic pressures for the given solutions are: - a. 48.02 atm - b. 6.67 atm - c. 52.16 atm - d. 45.57 atm

Key Concepts

Van't Hoff Factor

The van’t Hoff factor is an important concept when calculating the osmotic pressure for solutions. It represents the number of ions or particles that are generated when a solute dissolves. In simpler terms, it tells us how many pieces a compound breaks into once it is in a solution.
For non-electrolytes, which do not dissociate into ions and remain intact as molecules, the van't Hoff factor, represented by the letter \(i\), is 1. Examples include methanol and glycerol, both of which are non-electrolytes and therefore have \(i = 1\).
In contrast, electrolytes do dissociate into ions in solutions. For instance, \(\text{MgCl}_2\) breaks into three ions: one \(\text{Mg}^{2+}\) and two \(\text{Cl}^-\) ions, thus having a van't Hoff factor \(i = 3\). Similarly, \(\text{CaCl}_2\) also dissociates into three ions: one \(\text{Ca}^{2+}\) and two \(\text{Cl}^-\), therefore \(i = 3\). Understanding the van't Hoff factor is crucial in calculating accurate osmotic pressure as it adjusts the concentration of particles in the solution.

Non-Electrolyte

Non-electrolytes are substances that do not separate into ions when they are in solution. Instead, they dissolve as whole molecules, which means they do not contribute to electrical conductivity in the solution.
Examples of non-electrolytes include organic compounds like methanol and glycerol. In water, these compounds simply mix with the water molecules, maintaining their molecular integrity.
Because non-electrolytes do not produce ions, their van’t Hoff factor \(i\) is always equal to 1. This characteristic simplifies the calculation of their osmotic pressure since there is no need to adjust the concentration for ion dissociation. Understanding the behavior of non-electrolytes helps in predicting how they will affect properties like osmotic pressure in a solution.

Molar Concentration

Molar concentration, often denoted by \(M\), is a way of expressing the amount of solute present in a given volume of solution. It is an important measurement used in chemistry to define the concentration of a substance in a solution.
The molar concentration is calculated by dividing the number of moles of solute by the volume of the solution in liters.
For example, if we have 0.549 moles of glycerol in a solution with a total volume of 0.250 liters, the molar concentration is \(\frac{0.549}{0.250} = 2.196 \text{ M}\).
Knowing the molar concentration is essential in osmotic pressure calculations, as it directly affects how concentrated the solution is. Higher molar concentrations generally lead to higher osmotic pressures when calculated using the formula for osmotic pressure.

Conversion to Kelvin

Temperature plays a key role in many scientific calculations, including those involving osmotic pressure. To ensure accuracy, temperatures must often be converted into Kelvin, the SI unit of temperature. This is because scientific formulas are typically defined using Kelvin.
To convert Celsius to Kelvin, we simply add 273.15 to the Celsius temperature. For instance, a temperature of 20°C becomes \(20 + 273.15 = 293.15 \text{ K}\).
Using Kelvin in calculations ensures uniformity and consistency across different regions and systems. This conversion is crucial in chemistries involving gases and solutions, where temperature impacts the behavior of particles and their interactions.