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A common practice in cooking is the addition of salt to boiling water (Kb = 0.52 °C kg/mole). One of the reasons for this might be to raise the temperature of the boiling water. If 2.85 kg of water is boiling at 100 °C, how much NaCl (MW = 58.44 g/mole) would need to be added to the water to increase the boiling point by 2 °C? Must show your work.

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Answer:

640.59 g of NaCl.

Step-by-step explanation:

Using Boiling point elevation,

ΔTb = kb * i * b

Where,

b = molality

kb = 0.52 °C kg/mole

i = van't koff factor

ΔTb = ΔTsol - ΔTsolv

= 2°C

2 = 0.52 * b

= 3.846 mol/kg.

Mass of water, solvent = 2.85 kg

Molality is the ratio of the number of moles of the solute to the mass of the solvent.

= 3.846 * 2.85

= 10.96 mol.

Mass of NaCl = molar mass * number of moles

= 58.44 * 10.96

= 640.59 g of NaCl.

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