Question

Two elements A and B form compounds having molecular formula \(\mathrm{AB}_{2}\) and \(\mathrm{AB}_{4}\). When dissolved in \(20 \mathrm{~g}\) of \(\mathrm{C}_{6} \mathrm{H}_{6}, 1 \mathrm{~g}\) of \(\mathrm{AB}_{2}\) lowers the freezing point by \(2.55 \mathrm{~K}\), whereas \(1.0 \mathrm{~g}\) of \(\mathrm{AB}_{4}\) lowers it by \(1.7 \mathrm{~K}\). The molar depression constant for benzene is \(5.1 \mathrm{~K}-\mathrm{kg} \mathrm{mol}^{-1}\). The atomic masses of \(\mathrm{A}\) and \(\mathrm{B}\) are (a) 50,25 (b) 50,50 (c) 25,50 (d) 75,25

Short answer

The atomic masses of A and B are 50 and 25, respectively (option a).

Step-by-step solution

Calculate Molality for AB2 and AB4

Identify and use the relation that freezing point depression, DeltaTf, is the product of the molality (m), the molal depression constant (Kf), and the van't Hoff factor (i). For AB2, DeltaTf_AB2 = Kf * m_AB2 * i_AB2. For AB4, DeltaTf_AB4 = Kf * m_AB4 * i_AB4. Calculate molality (m) for both AB2 and AB4 using DeltaTf_AB2 = 2.55 K and DeltaTf_AB4 = 1.7 K and given Kf = 5.1 K*kg/mol.

Calculate the van't Hoff Factor for AB2 and AB4

Solve for the van't Hoff factor (i) for both compounds. Since 1 gram of each substance is used, and given that solvents have the same mass, the molality is also the moles per kilogram of solvent. Use the formula i = DeltaTf / (m * Kf).

Determine Molar Masses of AB2 and AB4

From the steps above, calculate the molar mass of both AB2 and AB4 using their respective molalities and the masses of the solute (1 gram for each), as Molar Mass = Mass of Solute / (Molality * Mass of Solvent).

Apply the Law of Multiple Proportions

According to the law of multiple proportions, if A and B combine to form two different compounds (AB2 and AB4 in this case), the ratio of the masses of B that combine with a fixed mass of A should be a simple whole number. Establish the ratio of molar masses of AB2 and AB4 and find a whole number ratio for B in AB2 and AB4.

Set Up Equations for Atomic Masses

Using the whole number ratio from Step 4, set up two equations assuming atomic masses for A and B. One equation will be for the molar mass of AB2 and the other for AB4. Solve these simultaneous equations for the atomic masses of A and B.

Solve the Equations and Find Atomic Masses

Solve the simultaneous equations obtained in Step 5 to find the values of the atomic masses of A and B. Verify these values against the options given to find the correct answer.

Key concepts

Freezing Point Depression

Freezing point depression is a colligative property, meaning it depends on the quantity of solute particles in a solvent, not on the type of particles. When a solute is dissolved in a solvent, it usually lowers the freezing point of the solvent. This is because the solute particles disrupt the formation of the crystal lattice necessary for the solvent to solidify, thus requiring a lower temperature to freeze.

In the given exercise, the substances AB2 and AB4, when dissolved in benzene, cause the freezing point of benzene to decrease by 2.55 K and 1.7 K, respectively. This demonstrates freezing point depression, which is crucial when calculating the molecular properties of an unknown compound.

Molality

Molality is a measure of the concentration of a solution, defined as the number of moles of solute per kilogram of solvent. It is a useful unit in colligative property equations because it doesn’t vary with temperature, unlike molarity which is volume-dependent.

In our problem, molality plays a key role because we use it alongside the freezing point depression data and the molar depression constant to calculate the molar masses of AB2 and AB4. The connection between molality and freezing point depression is captured in the formula:
\[ \Delta T_f = K_f \times m \times i \], where \( \Delta T_f \) is the freezing point depression, \( K_f \) is the molar depression constant of the solvent, \( m \) is the molality, and \( i \) is the van't Hoff factor.

van't Hoff Factor

The van't Hoff factor (i) is a dimensionless parameter that represents the number of particles a solute yields when it dissolves in a solvent. It influences colligative properties such as boiling point elevation and freezing point depression. For nonelectrolytes like AB2 and AB4, which do not dissociate into ions, the van't Hoff factor is typically close to 1.

In the solution steps, we calculate the van't Hoff factor using the measured freezing point depression and molality. The van't Hoff factor, in turn, helps in evaluating the molarity and types of particles in a solution, which is essential in determining molar mass and ultimately helps in identifying unknown substances.

Law of Multiple Proportions

This law states that when two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in simple whole number ratios. It is foundational in stoichiometry and aids in determining the formulas of chemical compounds.

When applying the law of multiple proportions to the compounds AB2 and AB4, it indicates that the mass ratios of B in these compounds with a fixed mass of A are in a simple whole number ratio. Determining this ratio helps guide us towards understanding the stoichiometry of the compounds, which is critical for finding the actual atomic masses of elements A and B.

Molar Mass Calculation

Molar mass is the mass in grams of one mole of a substance. It plays a pivotal role in chemistry for converting between the mass of a substance and the number of moles. Calculating the molar mass involves dividing the mass of the solute by the product of its molality and the solvent's mass in kilograms.

For the compounds in the exercise, AB2 and AB4, we utilize the calculated molality and apply the relationship:
\[ \text{Molar Mass} = \frac{\text{Mass of Solute}}{\text{Molality} \times \text{Mass of Solvent}} \]. This formula, combined with the colligative property of freezing point depression and the law of multiple proportions, allows us to deduce the molar masses of AB2 and AB4, as well as to estimate the atomic masses of A and B.