Question

A \(40.0 \%\) by weight solution of \(\mathrm{KSCN}\) in water at \(20^{\circ} \mathrm{C}\) has a density of \(1.22 \mathrm{~g} / \mathrm{mL}\). (a) What is the mole fraction of KSCN in the solution, and what are the molarity and molality? (b) Given the calculated mole fraction of salt in the solution, comment on the total number of water molecules available to hydrate each anion and cation. What ion pairing (if any) would you expect to find in the solution? Would you expect the colligative properties of such a solution to be those predicted by the formulas given in this chapter? Explain.

Short answer

The solution is 40.0% KSCN by weight, which means there are 40.0 g of KSCN and 60.0 g of water. We'll find the number of moles of each substance and calculate the mole fraction of KSCN. Then, we can find the molarity and molality of KSCN in the solution. Given the mole fraction of KSCN, we'll determine the number of water molecules per ion and discuss possible ion pairing and the expected colligative properties of the solution. Based on our calculations, the mole fraction of KSCN is \(\chi_{KSCN} = 0.0946\), the molarity (M) is \(1.61\,M\), and the molality (m) is \(2.97\,mol/kg\). There are approximately 10 water molecules per ion in the solution and ion pairing is likely in a concentrated solution like this. The colligative properties of such a solution might be different from those predicted by the formulas in the chapter due to the presence of ion pairing.

Step-by-step solution

Convert given weight percentages to mass

The solution is 40.0% KSCN by weight, which means there are 40.0 g of KSCN in 100 g of the solution, and the remaining part, which is water, weighs 60.0 g.

Calculate the number of moles of KSCN and water

We'll use the molar masses to find the number of moles of each substance. \(M_{KSCN} = 39.1 + 32.1 + 14.0 = 85.2\,g/mol \newline M_{H_2O} = 18.0\,g/mol \newline n_{KSCN} = \frac{40.0 \,g}{85.2\,g/mol} \newline n_{H_2O} = \frac{60.0\,g}{18.0\,g/mol}\)

Calculate the mole fraction of KSCN

We can calculate the mole fraction of KSCN using the formula: \(\chi_{KSCN} = \frac{n_{KSCN}}{n_{KSCN} + n_{H_2O}}\)

Calculate the molarity and molality of KSCN

Using the density provided, we can calculate the volume of 100 g of the solution: \(V = \frac{m_{solution}}{\rho} = \frac{100\,g}{1.22\,g/mL}\) Now, we can find the molarity (M) of KSCN: \(M = \frac{n_{KSCN}}{V}\) To calculate the molality (m), we divide the number of moles of KSCN by the mass of water in kilograms: \(m = \frac{n_{KSCN}}{m_{H_2O} (kg)}\)

Discuss the hydration and possible ion pairing

Given the mole fraction of KSCN calculated in Step 3, we determine the number of water molecules per ion in the solution. Based on this information, we can discuss the possible ion pairing and the expected behavior of the solution.

Discuss the colligative properties

Comment on whether the colligative properties of this solution would be as predicted by the formulas in the chapter, based on the possible ion pairing and the number of water molecules available for hydration.

Key concepts

Mole Fraction

The concept of mole fraction is a way of expressing the concentration of a constituent in a mixture. It is defined as the ratio of moles of a component to the total moles of all components in the solution. This method is particularly useful in understanding the composition of chemical systems, especially when dealing with non-ideal solutions. For the given exercise, the mole fraction of KSCN is calculated using the formula:

• The number of moles of KSCN and water are deduced from their respective masses and molar masses.
• The mole fraction of KSCN is \[\chi_{KSCN} = \frac{n_{KSCN}}{n_{KSCN} + n_{H_2O}}\]

The mole fraction gives insight into the ratio of solute to solvent molecules, impacting properties like vapor pressure and boiling point. Understanding mole fractions is crucial in predicting the behavior of chemical solutions.

Molarity

Molarity is a common and very practical expression of solution concentration, defined as the number of moles of solute per liter of solution. It is denoted by the symbol \(M\). For the calculation in the exercise:

• Molarity is obtained by dividing the calculated moles of KSCN by the total volume of the solution (in liters) derived from its mass and density, \[M = \frac{n_{KSCN}}{V}\].
• Volume is calculated using the formula \(V = \frac{m_{solution}}{\rho}\), where \(\rho\) is the density.

Molarity is widely used in laboratories because it provides a straightforward concentration metric when preparing chemical reactions, allowing chemists to accurately measure and mix reactants. It's essential in stoichiometry calculations in chemical equations.

Molality

Molality is another way to express solution concentration, similar to molarity but using mass instead of volume. It is defined as moles of solute per kilogram of solvent. This concept avoids temperature fluctuations since masses do not change with temperature. The exercise demonstrates how to compute molality:

• Molality is determined by dividing the moles of KSCN by the mass of water (in kilograms), \[m = \frac{n_{KSCN}}{m_{H_2O} (kg)}\].

Molality is particularly useful when dealing with colligative properties, which depend on the number of particles in a solution rather than their identity. Understanding molality helps in fields where precise concentrations are necessary, such as pharmacology or chemical engineering.

Colligative Properties

Colligative properties are characteristics of solutions that depend on the ratio of solute particles to solvent molecules, rather than the identity of the solute. These properties include boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure. Here's why they're important:

• Colligative properties arise because the addition of solute particles changes the physical properties of the solvent.
• For the exercise, knowing the mole fraction, molarity, and molality gives insight into the potential changes in these properties.
• Evaluating potential ion pairing and the availability of water molecules for hydration helps predict deviations from ideal behavior and effectiveness of colligative property formulas.

Mastering colligative properties allows one to explain why certain solutions behave differently despite having similar compositions and can be crucial in industrial applications such as antifreeze formulation and food preservation.