Question

A 0.500 -g sample of a compound is dissolved in enough water to form \(100.0 \mathrm{mL}\) of solution. This solution has an osmotic pressure of 2.50 atm at \(25^{\circ} \mathrm{C}\). If each molecule of the solute dissociates into two particles (in this solvent), what is the molar mass of this solute?

Short answer

The molar mass of the solute compound is approximately \(97.8 \frac{\mathrm{g}}{\mathrm{mol}}\).

Step-by-step solution

Convert the Temperature to Kelvin

To work with the osmotic pressure formula, we need to convert the given temperature from Celsius to Kelvin: \[ T_{K} = T_{C} + 273.15\] \[T_{K} = 25^{\circ}C + 273.15 = 298.15\,\mathrm{K}\]

Calculate the Moles of Solute Particles

Using the osmotic pressure formula, we can rearrange it to solve for the moles of solute particles: \[\cfrac{n}{V} = \cfrac{\Pi}{R \cdot T}\] We are given the osmotic pressure \(\Pi = 2.50\,\mathrm{atm}\), the volume of the solution \(V = 100.0\,\mathrm{mL} = 0.100\,\mathrm{L}\), and the ideal gas constant \(R = 0.0821\,\cfrac{\mathrm{L}\cdot\mathrm{atm}}{\mathrm{mol}\cdot\mathrm{K}}\). Now we can plug the values into the formula: \[\cfrac{n}{0.100\,\mathrm{L}} = \cfrac{2.50\,\mathrm{atm}}{(0.0821\,\cfrac{\mathrm{L}\cdot\mathrm{atm}}{\mathrm{mol}\cdot\mathrm{K}}) \cdot (298.15\,\mathrm{K})}\] From this, we can solve for the moles of solute particles, \(n\): \[n = 0.100\,\mathrm{L} \cdot \cfrac{2.50\,\mathrm{atm}}{(0.0821\,\cfrac{\mathrm{L}\cdot\mathrm{atm}}{\mathrm{mol}\cdot\mathrm{K}}) \cdot (298.15\,\mathrm{K})} = 0.0102\,\mathrm{mol}\]

Find the Moles of the Solute Compound

Since each molecule of the solute dissociates into two particles in the solvent, there are half as many moles of the solute compound as there are solute particles: \[n_{\mathrm{solute}} = \cfrac{n_{\mathrm{particles}}}{2}\] \[n_{\mathrm{solute}} = \cfrac{0.0102\,\mathrm{mol}}{2} = 0.0051\,\mathrm{mol}\]

Calculate the Molar Mass of the Solute

We are given the sample mass as 0.500 g. Now we can find the molar mass of the solute using the following equation: \[M_{\mathrm{solute}} = \cfrac{m_{\mathrm{solute}}}{n_{\mathrm{solute}}}\] \[M_{\mathrm{solute}} = \cfrac{0.500\,\mathrm{g}}{0.0051\,\mathrm{mol}} = 97.8\,\cfrac{\mathrm{g}}{\mathrm{mol}}\] The molar mass of the solute compound is approximately 97.8 g/mol.

Key concepts

Molar Mass Calculation

Molar mass is a fundamental concept in chemistry that refers to the mass of one mole of a substance, typically expressed in grams per mole. In this particular exercise, calculating the molar mass involves both understanding and applying the relationship between mass, moles, and the substance's property in solution.
The given problem states that a 0.500 g sample of a solute is dissolved. A crucial step in finding the molar mass is determining the number of moles of the solute. Using the information provided, we can substitute into the formula:

• Mass of the solute is 0.500 g.
• Moles of solute are calculated to be 0.0051 mol (after considering dissociation).

The key formula used is:\[ M_{\mathrm{solute}} = \frac{m_{\mathrm{solute}}}{n_{\mathrm{solute}}} \]By substituting the values, you achieve \[ M_{\mathrm{solute}} = \frac{0.500\,\mathrm{g}}{0.0051\,\mathrm{mol}} \approx 97.8\,\frac{\mathrm{g}}{\mathrm{mol}} \].
Molar mass helps determine the molecular weight of the solute in the solution and aids in identifying the solute's nature.

Dissociation in Solution

In many solutions, particularly with ionic compounds, dissociation occurs where molecules split into smaller particles, typically ions. The exercise notes that the solute dissociates into two particles. This dissociation is crucial because it affects the osmotic pressure and the calculation of moles.
Let's break down what happens when a substance dissociates:

• The given solute splits into two solute particles upon being dissolved.
• What it implies for our solution is that for every 1 mole of solute, there are 2 moles of solute particles contributing to osmotic pressure.

The formula used to find moles post-dissociation is:\[ n_{\mathrm{solute}} = \frac{n_{\mathrm{particles}}}{2} \]Where \( n_{\mathrm{particles}} \), derived from the osmotic pressure calculation, represents the molar quantity factoring in both particles.
Understanding dissociation is key to correlating the physical observations such as osmotic pressure to the chemical nature of the solute in solution.

Ideal Gas Constant

The ideal gas constant, denoted as \( R \), plays an essential role in the formula for calculating osmotic pressure. It appears in the pressure equation:\[ \pi = \frac{n}{V}RT \]where \( \pi \) is the osmotic pressure, \( n \) is the number of moles, \( V \) is the volume in liters, and \( T \) is the temperature in Kelvin.
In this exercise, the ideal gas constant \( R = 0.0821\,\frac{\mathrm{L} \cdot \mathrm{atm}}{\mathrm{mol} \cdot \mathrm{K}} \) is used for calculations in atmospheres and liters. This constant is crucial as it unifies the conditions of pressure, temperature, and volume.

• This constant helps relate osmotic pressure to concentration at a specific temperature.
• Note that the correct units for the constant, pressure, volume, and temperature must be followed to ensure accuracy in the calculated results.

Being aware of these relationships helps in understanding how solute properties affect osmotic behavior in solutions.