Question

Point 1 molal aqueous solution of lysine the ph at which no charge

Answer 1

∆Tb = Tb -Tbo = i.Kb.m

AB3 A3+ + 3B-

n = 4

a = i-1/n-1

0.9 = i-1/4-1

i = 3.7

Tbo = 373K

Thus,

Tb = (3.7*0.52*0.1)+373 = 373.192K

Answer 2

Your answer - :

∆Tb = Tb -Tbo = i.Kb.m

∆Tb = Tb -Tbo = i.Kb.mAB3 A3+ + 3B-

∆Tb = Tb -Tbo = i.Kb.mAB3 A3+ + 3B- n = 4

∆Tb = Tb -Tbo = i.Kb.mAB3 A3+ + 3B- n = 4

∆Tb = Tb -Tbo = i.Kb.mAB3 A3+ + 3B- n = 4 a = i-1/n-1

∆Tb = Tb -Tbo = i.Kb.mAB3 A3+ + 3B- n = 4 a = i-1/n-10.9 = i-1/4-1

∆Tb = Tb -Tbo = i.Kb.mAB3 A3+ + 3B- n = 4 a = i-1/n-10.9 = i-1/4-1i = 3.7

∆Tb = Tb -Tbo = i.Kb.mAB3 A3+ + 3B- n = 4 a = i-1/n-10.9 = i-1/4-1i = 3.7

∆Tb = Tb -Tbo = i.Kb.mAB3 A3+ + 3B- n = 4 a = i-1/n-10.9 = i-1/4-1i = 3.7 Tbo = 373K

∆Tb = Tb -Tbo = i.Kb.mAB3 A3+ + 3B- n = 4 a = i-1/n-10.9 = i-1/4-1i = 3.7 Tbo = 373K

∆Tb = Tb -Tbo = i.Kb.mAB3 A3+ + 3B- n = 4 a = i-1/n-10.9 = i-1/4-1i = 3.7 Tbo = 373K Thus,

∆Tb = Tb -Tbo = i.Kb.mAB3 A3+ + 3B- n = 4 a = i-1/n-10.9 = i-1/4-1i = 3.7 Tbo = 373K Thus,Tb = (3.7*0.52*0.1)+373 = 373.192K