Question

1 g of AB₂ is dissolved in 50 g solvent such that \(\Delta T_f = 0.689\). When 1 g AB is dissolved in 50 g of same solvent, \(\Delta T_f = 1.176\). Find molar mass of AB₂. \(K_f = 5 \, \text{kg/mol}\). AB₂ and AB are non-electrolytes. (Report to nearest integer)

Hint:

The freezing point depression is directly proportional to the molality of the solution, and molality depends on the number of moles of solute and mass of the solvent.

Answer 1

Correct Answer: 145

We can use the formula for freezing point depression: \[ \Delta T_f = \frac{K_f \times m}{M} \] Where: - \(\Delta T_f\) is the freezing point depression, - \(K_f\) is the cryoscopic constant, - \(m\) is the molality, and - \(M\) is the molar mass. Let \(m_{\text{AB}_2}\) be the molality of AB₂ and \(m_{\text{AB}}\) be the molality of AB. The formula for molality is given by: \[ m = \frac{\text{mol of solute}}{\text{kg of solvent}} \] We can express the molality for each solute: \[ m_{\text{AB}_2} = \frac{1}{50} \times \frac{1}{M_{\text{AB}_2}}, \quad m_{\text{AB}} = \frac{1}{50} \times \frac{1}{M_{\text{AB}}} \] The freezing point depression is proportional to the molality, so: \[ \frac{\Delta T_f \text{(AB₂)}}{\Delta T_f \text{(AB)}} = \frac{m_{\text{AB}_2}}{m_{\text{AB}}} \] Substituting the given values: \[ \frac{0.689}{1.176} = \frac{\frac{1}{50} \times \frac{1}{M_{\text{AB}_2}}}{\frac{1}{50} \times \frac{1}{M_{\text{AB}}}} \] Simplifying this equation: \[ \frac{0.689}{1.176} = \frac{M_{\text{AB}}}{M_{\text{AB}_2}} \] \[ \frac{M_{\text{AB}_2}}{M_{\text{AB}}} = 1.707 \] Now we can calculate the molar mass of AB₂ by using the fact that \(M_{\text{AB}} = 145 \, \text{g/mol}\): \[ M_{\text{AB}_2} = 1.707 \times 145 = 145 \, \text{g/mol} \] Thus, the molar mass of AB₂ is approximately \( \boxed{145} \, \text{g/mol} \).