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
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.
is the Kb 0.51 just standard BP of water? i think that's where i'm getting lost.
To calculate the boiling point of each solution, we need to know the molality (m) and the boiling point elevation constant (Kb) for each solute. The equation you mentioned, dTb = Kb * m, is correct.
However, in this case, the question did not provide the boiling point elevation constant (Kb) for either HCl or C6H12O6. Therefore, it is not possible to directly determine the boiling point elevation using the provided information.
The boiling point elevation constant (Kb) is a property specific to each solute and solvent pair, and it represents the degree to which the boiling point of the solvent increases due to the addition of the solute. Without this information, we cannot calculate the boiling point elevation.
To find the boiling point elevation constant (Kb) for a given solute and solvent, you would typically need to consult a reference table or use experimental data. So, it is essential to have this information to calculate the boiling point elevation accurately.