Problem 105 - Chemistry The Central Science In Si Units

Question

A lithium salt used in lubricating grease has the formula \(\mathrm{LiC}_{n} \mathrm{H}_{2 n+1} \mathrm{O}_{2}\). The salt is soluble in water to the extent of \(0.036 \mathrm{~g}\) per \(100 \mathrm{~g}\) of water at \(25^{\circ} \mathrm{C}\). The osmotic pressure of this solution is found to be \(7.61 \mathrm{kPa}\). Assuming that molality and molarity in such a dilute solution are the same and that the lithium salt is completely dissociated in the solution, determine an appropriate value of \(n\) in the formula for the salt.

Step-by-Step Solution

Answer

The value of \(n\) for the lithium salt is approximately 16.

Step 1: Understand the Problem

We are given a lithium salt with the formula \(\mathrm{LiC}_n\mathrm{H}_{2n+1}\mathrm{O}_2\), which is soluble in water. The problem provides the solubility and osmotic pressure at \(25^\circ \text{C}\). Our task is to find the value of \(n\) in the formula.

Step 2: Use Osmotic Pressure Formula

The osmotic pressure \(\pi\) is given by the equation \(\pi = iMRT\), where \(i\) is the van't Hoff factor (\(i = 2\) for complete dissociation), \(M\) is the molarity of the solution, \(R\) is the universal gas constant (\(8.314 \; \text{J/mol·K}\)), and \(T\) is the temperature in Kelvin (\(T = 298 \; \text{K}\)). Thus, \(7.61 = 2 \times M \times 8.314 \times 298\).

Step 3: Calculate Molarity

Rearrange the formula to solve for molarity \(M\): \[ M = \frac{7.61}{2 \times 8.314 \times 298} \approx 0.00153 \; \text{mol/L}\].

Step 4: Relate Molarity to Mass and Molar Mass

We know that \(0.036 \; \text{g}\) of the salt dissolve in \(100 \; \text{g}\) of water. So, in \(1 \; \text{L} \approx 1000 \; \text{g}\) of water, the mass dissolved would be \(0.36 \; \text{g}\). Using \(M = \frac{\text{mass}}{\text{molar mass} \times \text{volume}}\), and knowing \(M \approx 0.00153 \; \text{mol/L}\), find the molar mass by rearranging as \(\text{molar mass} = \frac{0.36}{0.00153} \approx 235.29 \; \text{g/mol}\).

Step 5: Determine Molar Mass and Find n

The molar mass of \(\mathrm{LiC}_n\mathrm{H}_{2n+1}\mathrm{O}_2\) can be determined by adding the atomic masses: Li (approx. 6.94), C (12.01 each), H (1.008 each), and O (16.00 each). Assemble the equation: \[6.94 + 12.01n + 1.008(2n+1) + 2 \times 16.00 = 235.29\]. Solve this for \(n\) using algebra to simplify and find approximately \(n = 16\).

Key Concepts

Lithium Salt

Lithium salts are chemical compounds composed of lithium ions and a salt component. Here, we are dealing with a lithium salt formula of \(\text{LiC}_n\text{H}_{2n+1}\text{O}_2\). Lithium salts have a wide range of applications, notably in lubricating grease, where they act to lower friction and wear between surfaces.

In solutions, lithium salts often dissociate completely, releasing lithium ions. This property is critical when predicting behaviors in osmotic pressure scenarios, as the dissociation impacts the van't Hoff factor \(i\) used in calculations. In our problem, the lithium salt dissociates to provide an \(i\) value of 2. This understanding helps us to effectively compute the molarity necessary to determine other characteristics such as molar mass and osmotic pressure.

Molarity

Molarity is a crucial concept when solving problems related to solutions, as it measures the concentration of a solute in a solution. Molarity \(M\) is defined as the number of moles of solute per liter of solution. It is expressed as \(M = \frac{n}{V}\), where \(n\) is the moles of solute, and \(V\) is the volume of the solution in liters.

In the given problem, we use molarity to find out the concentration of the lithium salt in solution. Given the osmotic pressure and assuming the solution is dilute, we applied the formula for osmotic pressure \(\pi = iMRT\) to find \(M\). With known values of \(\pi\), \(i\), \(R\), and \(T\), we determined the molarity to be approximately 0.00153 mol/L. This value played a pivotal role in calculating the molar mass needed to find \(n\) in the chemical formula.

Molar Mass

Molar mass is another key concept in chemistry, representing the mass of a given substance (chemical element or chemical compound) divided by the amount of substance (mole). It is expressed in g/mol. Molar mass is calculated by summing the atomic masses of all atoms in a molecule.

For the lithium salt, \(\text{LiC}_n\text{H}_{2n+1}\text{O}_2\), the molar mass was crucial in deducing the value of \(n\). First, we found the molar mass by dividing the mass of the solute dissolved in a certain volume of water by the molarity, \(M\). This gives a calculated molar mass of \(235.29 \; \text{g/mol}\).

To determine \(n\), we constructed an equation using the atomic masses of lithium, carbon, hydrogen, and oxygen. Solving this equation gave us an approximate \(n\) value of 16, which finalized our solution.

Dilute Solution

A dilute solution is one where a relatively small amount of solute is dissolved in a solvent. Such solutions are often considered simplistically in calculations, assuming negligible solute-solvent interactions affecting properties like volume.

In the exercise, the given lithium salt solution is dilute, and this allowed us to assume that molality and molarity could be considered equivalent. This simplification was key in successfully applying the osmotic pressure formula.

Dilute solutions make it easier to assume ideal behavior and simplify equations, focusing only on essential interactions. This assumption allowed us to accurately compute other solution properties without dealing with complex interactions affecting solution behavior.