Two elements 'A' and 'B' form compounds having formulae `AB_(2)` and `AB_(4)`. When dissolved in 20g of benzene, 1g of `AB_(2)` lowers the freezing point by `2.3^(@)C` whereas `1g` of `AB_(4)` lowers the freezing point by `1.3^(@)C`. Calculate the atmoic weights of A and B . (`K_(f)` for benzene `=5.1^(@)Cm^(-1)`)

Answer

To solve the problem, we need to calculate the atomic weights of elements A and B based on the freezing point depression caused by their compounds \( AB_2 \) and \( AB_4 \) when dissolved in benzene. We will use the formula for freezing point depression:

\[ \Delta T_F = K_F \cdot m \]

where:
- \( \Delta T_F \) is the depression in freezing point,
- \( K_F \) is the cryoscopic constant of the solvent (benzene in this case),
- \( m \) is the molality of the solution.

### Step 1: Calculate the molality for \( AB_2 \)

Given:
- \( \Delta T_F \) for \( AB_2 = 2.3^\circ C \)
- Mass of \( AB_2 = 1 \, g \)
- Mass of solvent (benzene) = 20 g
- \( K_F \) for benzene = 5.1 \( ^\circ C \cdot m^{-1} \)

Using the freezing point depression formula, we can rearrange it to find the molality \( m \):

\[ m = \frac{\Delta T_F}{K_F} \]

Substituting the values:

\[ m_{AB_2} = \frac{2.3}{5.1} = 0.45098 \, mol/kg \]

### Step 2: Calculate the number of moles of \( AB_2 \)

Since molality \( m \) is defined as moles of solute per kg of solvent, we can calculate the number of moles of \( AB_2 \):

\[ \text{Mass of solvent in kg} = \frac{20}{1000} = 0.02 \, kg \]

Using the molality:

\[ \text{Moles of } AB_2 = m \cdot \text{mass of solvent in kg} = 0.45098 \cdot 0.02 = 0.0090196 \, mol \]

### Step 3: Calculate the molar mass of \( AB_2 \)

The molar mass \( M_{AB_2} \) can be calculated using the formula:

\[ M_{AB_2} = \frac{\text{mass of solute}}{\text{moles of solute}} = \frac{1 \, g}{0.0090196 \, mol} \approx 110.87 \, g/mol \]

### Step 4: Calculate the molality for \( AB_4 \)

Now, for compound \( AB_4 \):

Given:
- \( \Delta T_F \) for \( AB_4 = 1.3^\circ C \)
- Mass of \( AB_4 = 1 \, g \)

Using the same formula for molality:

\[ m_{AB_4} = \frac{1.3}{5.1} = 0.25490 \, mol/kg \]

### Step 5: Calculate the number of moles of \( AB_4 \)

Using the mass of solvent (20 g = 0.02 kg):

\[ \text{Moles of } AB_4 = m \cdot \text{mass of solvent in kg} = 0.25490 \cdot 0.02 = 0.005098 \, mol \]

### Step 6: Calculate the molar mass of \( AB_4 \)

\[ M_{AB_4} = \frac{\text{mass of solute}}{\text{moles of solute}} = \frac{1 \, g}{0.005098 \, mol} \approx 196.15 \, g/mol \]

### Step 7: Set up equations for atomic weights of A and B

Let the atomic weights of A and B be \( M_A \) and \( M_B \) respectively. From the compounds:

1. For \( AB_2 \):
\[ M_A + 2M_B = 110.87 \quad \text{(Equation 1)} \]

2. For \( AB_4 \):
\[ M_A + 4M_B = 196.15 \quad \text{(Equation 2)} \]

### Step 8: Solve the equations

Subtract Equation 1 from Equation 2:

\[ (M_A + 4M_B) - (M_A + 2M_B) = 196.15 - 110.87 \]
\[ 2M_B = 85.28 \]
\[ M_B = 42.64 \, g/mol \]

Substituting \( M_B \) back into Equation 1:

\[ M_A + 2(42.64) = 110.87 \]
\[ M_A + 85.28 = 110.87 \]
\[ M_A = 110.87 - 85.28 = 25.59 \, g/mol \]

### Final Result

The atomic weights of elements A and B are:
- \( M_A \approx 25.6 \, g/mol \)
- \( M_B \approx 42.64 \, g/mol \)