Topscore chem problem regarding boiling point, solute, solvent

[Aschematic]

Advertisement - Members don't see this ad

If the Kb of a solvent is 3.51°C/m, and the boiling point of the pure solvent is 70.4°C, then what is the boiling point of a 4m solution of a solute dissolved in the solvent?

The answer is 84.4°C.

I don't have any background whatsoever on solutions, so I would very much appreciate it if you could please explain the process of getting this answer...

Thank you so much. Looking forward to replies.

 

[wantVCUdental]

If the Kb of a solvent is 3.51°C/m, and the boiling point of the pure solvent is 70.4°C, then what is the boiling point of a 4m solution of a solute dissolved in the solvent?

The answer is 84.4°C.

I don't have any background whatsoever on solutions, so I would very much appreciate it if you could please explain the process of getting this answer...

Thank you so much. Looking forward to replies.

Don't worry, just know how to calculate, remember these two equations:
boiling point elevation change in T=kb m i
freezing point depressing change in T=kf m i
in both equations, m is the molality, i is the dissocation factor (i.e NaCl becomes 2 ions, so i=2, glucose can't dissocate, so i=1)

In this Q, it doesn't specify what is dissolved, so you don't worry about i
take m=4, multiply by Kb=3.51
so the change in temp is 14.04 about 14
14+70.4=84.4

Hope that helps

 

[Aschematic]

Thanks, wantVCUdental.

That helped a lot. 🙂