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What is the boiling point elevation of a solution containing 1 mol of NaCl in 1
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)
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Q1
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)
0.512 °C
1.024 °C
1.536 °C
2.048 °C
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Boiling point elevation = i * K_b * m = 2 * 0.512 * 1 = 1.024 °C (i = 2 for NaCl)
Questions & Step-by-step Solutions
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Q
Q: 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)
Solution:
Boiling point elevation = i * K_b * m = 2 * 0.512 * 1 = 1.024 °C (i = 2 for NaCl)
Steps: 6
Show Steps
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.
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