Question

A $0.500-\mathrm{g}$ 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 disso- ciates into two particles (in this solvent), what is the molar mass of this solute?

Short answer

The molar mass of the solute is 98.0 g/mol. This is determined by first converting the temperature to Kelvin, calculating the molar concentration of the solute (considering the dissociation), finding the number of moles of solute, and finally dividing the mass of the sample by the number of moles.

Step-by-step solution

Convert temperature to Kelvin

To use the formula for osmotic pressure, we must first convert the temperature given in Celsius to Kelvin. The conversion is as follows: $K=°C+273.15$ Given T = 25°C: $T=25+273.15=298.15K$

Calculate the molar concentration

Now that we have the temperature in Kelvin, we can rearrange the osmotic pressure formula to find the molar concentration, c: $c=\frac{\pi }{RT}$ Given π = 2.50 atm, R = 0.0821 L·atm·K^(-1)·mol^(-1), and T = 298.15 K: $c=\frac{2.50}{(0.0821)(298.15)}=0.102\mathrm{mol}/\mathrm{L}$ This value represents the molar concentration of the solute particles after dissociation, which is two particles per molecule. So, the actual molar concentration of the solute is half of this value: $c_{solute}=\frac{c}{2}=0.051\mathrm{mol}/\mathrm{L}$

Find the number of moles of solute

Now that we have the molar concentration, we can use the volume of the solution (100 mL) to find the number of moles of solute: Moles of solute = Molar concentration of solute × Volume Given the volume = 100.0 mL, Moles of solute = 0.051 mol/L × 0.100 L = 0.0051 mol

Calculate the molar mass of the solute

Finally, we can now find the molar mass of the solute by dividing the sample's mass by the number of moles: Molar mass = Mass of sample / Moles of solute Given mass of sample = 0.500 g, Molar mass = $\frac{0.500\mathrm{g}}{0.0051\mathrm{mol}}$ = 98.0 $\mathrm{g}/\mathrm{mol}$ Thus, the molar mass of the solute is 98.0 g/mol.

Key concepts

Molar Mass

Understanding the molar mass of a substance is essential in chemistry, especially when dealing with solutions. The molar mass is the weight in grams of one mole of a substance, which is numerically equivalent to its molecular weight in atomic mass units (amu). To calculate it, you sum the atomic weights of all atoms in the molecule. For instance, water (H₂O) has a molar mass of approximately 18.02 g/mol, since each of the two hydrogen atoms contributes 1.01 g/mol, and the oxygen atom contributes around 16.00 g/mol.

In our exercise, we find the molar mass by dividing the given mass of the dissolved compound by the number of moles of solute. This value will assist us in identifying the compound or at least verifying its purity. Remember, the accuracy of your molar mass calculation can significantly impact the results of any further calculations that use this value, such as determination of solution concentration and calculation of osmotic pressure.

Colligative Properties

Colligative properties are physical properties of solutions that depend on the ratio of the number of solute particles to the number of solvent molecules in a solution, and not on the nature of the chemical species present. These properties include boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure. They are essential for understanding how solutes affect the physical properties of solvents.

In the problem at hand, osmotic pressure is the colligative property we focus on. Osmotic pressure is the pressure that must be applied to a solution to prevent the inward flow of water across a semipermeable membrane. It's determined by the concentration of solute particles and temperature, which is why a correct calculation of the molar concentration directly influences the computation of osmotic pressure.

van't Hoff factor

The van't Hoff factor (i) is a dimensionless quantity used in calculations involving colligative properties. It describes the number of particles that a compound dissociates into when it dissolves in a solvent. For example, a van't Hoff factor of 1 would signify a non-dissociating molecule like glucose, whereas a factor of 2 indicates a compound like NaCl, which dissociates into two particles: a sodium ion (Na⁺) and a chloride ion (Cl^–).

In our textbook exercise, the solute dissociates into two particles, so the van't Hoff factor is 2. This information is crucial because it means that the observed osmotic pressure is due to twice the number of particles in the solution than if the solute didn't dissociate. Therefore, the presence of a van't Hoff factor affects both the molar concentration and the molar mass calculations for the solute.

Solution Concentration

The concentration of a solution tells us how much solute is present in a given volume of solvent, and it is a measure of how 'dense' a solution is with solute particles. Common units of concentration include molarity (moles per liter), molality (moles per kilogram of solvent), and percent composition by mass or volume.

In terms of our exercise, the concentration of the solution is crucial for calculating both the osmotic pressure and molar mass. By dividing the osmotic pressure by the gas constant and the temperature in kelvins, we get the molarity of particles in the solution. However, because each particle dissociates, the actual concentration of the solute prior to dissociation is half of this value. Understanding the concept of solution concentration is critical when interpreting or predicting the behavior of solutions in various scientific and industrial processes.