Problem 84 - Chemistry The Science in Context

Question

Calculate the osmotic pressure of each of the following aqueous solutions at \(27^{\circ} \mathrm{C}\) a. \(10.0 \mathrm{g}\) of \(\mathrm{NaCl}\) in \(1.50 \mathrm{L}\) of solution b. \(10.0 \mathrm{mg} / \mathrm{L}\) of \(\mathrm{LiNO}_{3}\) c. \(0.222 M\) glucose d. \(0.00764 M \mathrm{K}_{2} \mathrm{SO}_{4}\)

Step-by-Step Solution

Answer

Question: Calculate the osmotic pressures of the following aqueous solutions at 27°C: a) 10.0 g NaCl in 1.50 L of solution b) 10.0 mg/L LiNO3 c) 0.222 M glucose d) 0.00764 M K2SO4 Answer: a) The osmotic pressure of the NaCl solution is 5.63 atm. b) The osmotic pressure of the LiNO3 solution is 0.00710 atm. c) The osmotic pressure of the glucose solution is 5.47 atm. d) The osmotic pressure of the K2SO4 solution is 0.560 atm.

Step 1: Determine the van 't Hoff factor of NaCl

NaCl dissociates into two ions in solution: Na+ and Cl-. Thus, its van 't Hoff factor, i, is 2.

Step 2: Calculate the molarity of the NaCl solution

To find the molarity of NaCl, first determine the moles of NaCl: Moles of NaCl = (mass of NaCl) / (molar mass of NaCl) = (10 g) / (58.44 g/mol) = 0.171 mol Now, calculate the molarity: M = (moles of NaCl) / (volume of solution in Liters) = (0.171 mol) / (1.50 L) = 0.114 M

Step 3: Convert temperature from Celsius to Kelvin

T(K) = T(°C) + 273.15 = 27°C + 273.15 = 300.15 K

Step 4: Calculate the osmotic pressure

Osmotic Pressure = i * M * R * T = 2 * 0.114 M * 0.0821 L atm / mol K * 300.15 K = 5.63 atm ##Solution b: Calculate the osmotic pressure of 10.0 mg/L LiNO3##

Step 1: Determine the van 't Hoff factor of LiNO3

LiNO3 dissociates into two ions in solution: Li+ and NO3-. Thus, its van 't Hoff factor, i, is 2.

Step 2: Calculate the molarity of the LiNO3 solution

First, convert the mass concentration to grams per liter: Mass concentration (g/L) = 10.0 mg/L * (1 g / 1000 mg) = 0.010 g/L Now, calculate the molarity: M = (mass concentration) / (molar mass of LiNO3) = (0.010 g/L) / (68.95 g/mol) = 0.000145 M

Step 3: Use the same temperature conversion as in Solution a

T(K) = 300.15 K

Step 4: Calculate the osmotic pressure

Osmotic Pressure = i * M * R * T = 2 * 0.000145 M * 0.0821 L atm / mol K * 300.15 K = 0.00710 atm ##Solution c: Calculate the osmotic pressure of 0.222 M glucose##

Step 1: Determine the van 't Hoff factor of Glucose

Glucose (C6H12O6) is a non-electrolyte and does not dissociate in solution. Thus, its van 't Hoff factor, i, is 1.

Step 2: The molarity is given: M = 0.222 M

Step 3: Use the same temperature conversion as in Solution a

Step 11: T(K) = 300.15 K

Step 4: Calculate the osmotic pressure

Step 12: Osmotic Pressure = i * M * R * T = 1 * 0.222 M * 0.0821 L atm / mol K * 300.15 K = 5.47 atm ##Solution d: Calculate the osmotic pressure of 0.00764 M K2SO4##

Step 1: Determine the van 't Hoff factor of K2SO4

Step 13: K2SO4 dissociates into three ions in solution: 2 K+ and 1 SO4^2-. Thus, its van 't Hoff factor, i, is 3.

Step 2: The molarity is given: M = 0.00764 M

Step 3: Use the same temperature conversion as in Solution a

T(K) = 300.15 K

Step 4: Calculate the osmotic pressure

Osmotic Pressure = i * M * R * T = 3 * 0.00764 M * 0.0821 L atm / mol K * 300.15 K = 0.560 atm

Key Concepts

van 't Hoff factor

The van 't Hoff factor, often represented as "i," is crucial when calculating properties like osmotic pressure in solutions. It shows how many particles a compound forms when it dissolves in water.

Understanding whether a substance ionizes in solution and to what extent, helps predict changes in colligative properties. For electrolytes like

• Sodium chloride (\(\mathrm{NaCl}\)): dissociates completely into \(\mathrm{Na^{+}}\) and \(\mathrm{Cl^{-}}\). Hence, its van 't Hoff factor is \(i = 2\).
• Lithium nitrate (\(\mathrm{LiNO}_{3}\)): dissociates into \(\mathrm{Li^{+}}\) and \(\mathrm{NO}_{3}^{-}\) ions, making \(i = 2\) as well.
• For non-electrolytes like glucose, which do not dissociate in solution, the van 't Hoff factor is \(i = 1\).
• Potassium sulfate (\(\mathrm{K}_{2}\mathrm{SO}_{4}\)): dissociates into two \(\mathrm{K^{+}}\) ions and one \(\mathrm{SO}_{4}^{2-}\), giving a van 't Hoff factor of \(i = 3\).

Accurate use of the van 't Hoff factor allows for precise calculations of osmotic pressure, affecting processes such as reverse osmosis or even biological cell behavior.

molarity

Molarity (M) is a measure of the concentration of a solute in a solution, expressed as moles of solute per liter of solution. This measurement is vital in stoichiometry and plays a crucial role in determining various chemical properties of a solution.

To calculate molarity:

• Determine the number of moles of the solute. This is done by dividing the mass of the solute by its molar mass. For instance, \( \mathrm{NaCl} \) has a molar mass of 58.44 g/mol. So, 10 g of \( \mathrm{NaCl} \) would be \( 0.171 \) moles.
• Divide the moles by the volume of the solution in liters. With \(1.50 \, \mathrm{L} \) of solution, the molarity for \( \mathrm{NaCl} \) would be \( 0.114 \, \mathrm{M} \).

Understanding molarity helps in calculating osmotic pressure, which is a colligative property dependent on solute concentration, not the type of solute.

temperature conversion

When dealing with temperature in scientific calculations, especially regarding gas laws and colligative properties like osmotic pressure, it's essential to use the Kelvin scale. The Kelvin scale provides a direct and absolute measure of thermal energy, removing the complications of negative temperatures.

To convert from Celsius to Kelvin, a simple addition is required:

• Add \(273.15\) to the Celsius temperature. For example, \(27^{\circ} \mathrm{C}\) becomes \(300.15 \mathrm{K}\).

Using Kelvin ensures consistency across calculations involving gas constants like \(R = 0.0821 \, \mathrm{L} \, \mathrm{atm} / \mathrm{mol} \, \mathrm{K} \), a fundamental part of the equation used to determine osmotic pressure.

aqueous solutions

An aqueous solution refers to a solution where water is the solvent. When substances dissolve in water, they can exhibit different behavior based on their chemical nature.

• Electrolytes: These are typically ionic compounds like \(\mathrm{NaCl}\), \(\mathrm{LiNO}_{3}\), and \(\mathrm{K}_{2}\mathrm{SO}_{4}\). They dissociate into ions in water, which enhances their ability to conduct electricity.
• Non-electrolytes: Substances like glucose dissolve in water but do not dissociate into ions. They do not conduct electricity in their aqueous form.

The behavior of a solute in an aqueous solution affects its colligative properties, such as boiling point elevation, freezing point depression, and, importantly, osmotic pressure. Understanding these behaviors helps in comprehending the manipulation and application of solutions in fields ranging from chemistry labs to biological systems.