What is the boiling point of a solution containing 0.5 mol of KCl in 1 kg of water? (K_b for water = 0.512 °C kg/mol)

1 question

1 item

Q: What is the boiling point of a solution containing 0.5 mol of KCl in 1 kg of water? (K_b for water = 0.512 °C kg/mol)

Solution: Boiling point elevation = i * K_b * m = 2 * 0.512 * 0.5 = 0.512 °C; Boiling point = 100 + 0.512 = 100.512 °C

Steps: 8

Step 1: Identify the number of moles of KCl in the solution, which is given as 0.5 mol.
Step 2: Determine the van 't Hoff factor (i) for KCl. Since KCl dissociates into K+ and Cl-, i = 2.
Step 3: Find the boiling point elevation constant (K_b) for water, which is given as 0.512 °C kg/mol.
Step 4: Calculate the molality (m) of the solution. Since we have 0.5 mol of KCl in 1 kg of water, m = 0.5 mol/kg.
Step 5: Use the formula for boiling point elevation: Boiling point elevation = i * K_b * m.
Step 6: Substitute the values into the formula: Boiling point elevation = 2 * 0.512 * 0.5.
Step 7: Calculate the boiling point elevation: 2 * 0.512 * 0.5 = 0.512 °C.
Step 8: Add the boiling point elevation to the normal boiling point of water (100 °C): 100 °C + 0.512 °C = 100.512 °C.