Calculate the boiling point of a solution prepared by adding 15.0 g of

1.	Calculate the boiling point of a solution prepared by adding 15.0 g of NacI to 250,.0 gof water. (Kb for water -0.512 K kg mol-1, Molar mass of NaCI 58.44 g, AssumeNaCI undergoes complete dissociation)
Answer» Moles of NaCl = 15/58.44 = 0.25 moles of NaCL Kg of solvent = 250/1000 Kg = 0.25 Kg of water Therefore, Molality: = 0.25/0.25 = 1 molal For water change in BP: ΔTb= Kb* m = 0.52 * 1 = 0.52 K Since water boils at 373.15 K at 1.013 bar pressure, therefore the boiling of solution will be 373.15+0.52 = 373.67 K

Calculate the boiling point of a solution prepared by adding 15.0 g of NacI to 250,.0 gof water. (Kb for water -0.512 K kg mol-1, Molar mass of NaCI 58.44 g, AssumeNaCI undergoes complete dissociation)

Answer»

Moles of NaCl = 15/58.44

= 0.25 moles of NaCL

Kg of solvent = 250/1000 Kg

= 0.25 Kg of water

Therefore, Molality:

= 0.25/0.25

= 1 molal

For water change in BP:

ΔTb= Kb* m

= 0.52 * 1

= 0.52 K

Since water boils at 373.15 K at 1.013 bar pressure, therefore the boiling of solution will be 373.15+0.52 = 373.67 K