Which of the following aqueous(water as a solvent) solutions would have the lowest boiling point if they are..
A. C12H22O11 with i=1
b.NaClwith i=2
c.CaCl2 with i=3
d.Ba(CLO3)2 with i=3
Answer:
Delta T = kb x i x m
kb for water = 0.512
a. delta T = 0.512 x 1 x 0.1 = 0.0512 °C
The boiling point of the solution will be 100 + 0.0512 = 100.05 °C ( lowest boiling point because i = 1)
b delta T = 0.512 x 2 x 0.1 = 0.102
the b.p. will be 100+0.102 = 100.102 °c
c. delta T = 0.512 x 3 x 0.1 = 0.154
the b.p. will be 100+ 0.104 = 100.104 °C
d. delta T = 0.512 x 3 x 0.1 = 0.154
the b.p. will be 100.154 °C
no need for such calculations as all r equimoloal aqueous solutions.
delta T = i. kb.m
now, m and kb r same for all solutions. so we get,
delta T is directly proportional to i that is van't hoff factor.
since `i' is minimum for aqueous solution of sugar, delta T will be minimum. and if delta T will be minimum, elevation in boiling point is minimum. this implies that boiling point will not elevate much. this solution will have the lowest boiling point.
The answers post by the user, for information only, FunQA.com does not guarantee the right.
More Questions and Answers: