0.1m aqueous solution of HCl and 0.1m aqueous solution of C6H12O6.
calculate the boiling point of each
i know the eq. dTb=Kbm
but i have no temperatures. this is the question
4 answers
It is an aqueous solution. You know that the boiling point with no HCl is 100 C. Use Kb and the molarity to compute the change in B.P.
can you show me. i'm confused.
delta T = i*Kb*molality
i is the van't Hoff factor, which for HCl is 2 (it dissolves into two particles).
delta T = 2*0.512*0.1 = 0.102
Therefore, the new boiling point for water must be 100 C + 0.102 = ??
For the sugar solution,
i = 1 since it doesn't ionize.
delta T = i*Kb*m
delta T = 1*0.512*0.1 = 0.0512
New boiling point will be
100 C + 0.0512.
The problem is hoping you will see the difference between the two; i.e., the HCl ionizes to give two particles and has just twice the effect as the other one although the molality is the same.
i is the van't Hoff factor, which for HCl is 2 (it dissolves into two particles).
delta T = 2*0.512*0.1 = 0.102
Therefore, the new boiling point for water must be 100 C + 0.102 = ??
For the sugar solution,
i = 1 since it doesn't ionize.
delta T = i*Kb*m
delta T = 1*0.512*0.1 = 0.0512
New boiling point will be
100 C + 0.0512.
The problem is hoping you will see the difference between the two; i.e., the HCl ionizes to give two particles and has just twice the effect as the other one although the molality is the same.
is the Kb 0.51 just standard BP of water? i think that's where i'm getting lost.
1 answer