What is the boiling point elevation of a solution containing 1 mol of NaCl in 1

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

Options:

Correct Answer: 1.024 °C

Solution:

Boiling point elevation = i * K_b * m = 2 * 0.512 * 1 = 1.024 °C (i = 2 for NaCl)

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

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

Step 1: Identify the formula for boiling point elevation, which is ΔT_b = i * K_b * m.
Step 2: Determine the van 't Hoff factor (i) for NaCl. Since NaCl dissociates into 2 ions (Na+ and Cl-), i = 2.
Step 3: Identify the value of K_b for water, which is given as 0.512 °C kg/mol.
Step 4: Identify the molality (m) of the solution. Since there is 1 mol of NaCl in 1 kg of water, m = 1 mol/kg.
Step 5: Plug the values into the formula: ΔT_b = 2 * 0.512 * 1.
Step 6: Calculate the boiling point elevation: ΔT_b = 1.024 °C.

Boiling Point Elevation – The increase in the boiling point of a solvent when a solute is dissolved in it, calculated using the formula ΔT_b = i * K_b * m.
Van 't Hoff Factor (i) – A factor that accounts for the number of particles the solute dissociates into; for NaCl, it is 2 because it dissociates into Na+ and Cl-.
Molality (m) – A measure of the concentration of a solution, defined as the number of moles of solute per kilogram of solvent.