how many gram of urea should be dissolved in 400 gram water so that difference in boiling point and freezing point is 105 degree Celsius (Kb = 0.512, Kf = 1.86)
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Given, Tb - Tf = 105
since, Tb-Tbo = Kb.m
Tb = Kb.m+Tbo = 0.512.m+100
and
Tfo-Tf = Kf.m
Tf =-1.86.m
since Tbo is boiling point of water and Tfo is melting point of water
Substituting the values and solving for m we get
m = 2.11, where m is the molality,
m= no of moles of solute/mass of solvent in kg
mass of solvent = 0.4kg
no.of moles of urea = 0.844
Thus mass of urea = 0.844*60= 50.69 g
since, Tb-Tbo = Kb.m
Tb = Kb.m+Tbo = 0.512.m+100
and
Tfo-Tf = Kf.m
Tf =-1.86.m
since Tbo is boiling point of water and Tfo is melting point of water
Substituting the values and solving for m we get
m = 2.11, where m is the molality,
m= no of moles of solute/mass of solvent in kg
mass of solvent = 0.4kg
no.of moles of urea = 0.844
Thus mass of urea = 0.844*60= 50.69 g
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