Answers
Number 2:
Part a:
Total mass = 500 gm
Mass % of sodium carbonate = 30 %
Use:
Mass percent = mass of sodium carbonate *100 / Total mass
30 = mass of sodium carbonate *100 / 500
mass of sodium carbonate = 150 gm
Mass of water = 500-150 = 350 gm
Since density of water is 1gm/ml, Above solution can be prepared by dissolving 150 gm of sodium carbonate in 350 ml of water.
Part B:
Molar mass of sodium carbonate = 105.99 gm = 106 gm
mass of sodium carbonate = 150 gm
Mass of water/solvent = 350 gm = 0.35 Kg
Number of moles of sodium carbonate = mass of sodium carbonate /Molar mass of sodium carbonate
= 150/106
=1.415 mol
Molality = Number of moles of solute / mass of solvent in Kg
= 1.415 / 0.35
= 4.043 m
Part C:
depression in freezing point = Kbp * molality
=0.512*4.043
= 2.07 oC
New Freezing point = 0 - 2.07 oC
= -2.07 oC
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