[desertfox23]

Hi all, if you all don't mind, I have a couple of questions that I need some help with and would really appreciate some assistance.


Here is the first problem:

What mass percent of KBr in water would give you a -1.5^o freezing point? 


delta temp (-1.5) = (2 ions) (x molality) (-1.86)
               (-1.5) = -3.72 (x molality)
               (-1.5) / (-3.72) = (x molality)
                                        molality = .403 moles per kg


I don't know what to do next. KBr has an atomic weight of 119.0023

Am I supposed to multiply .403 by 119.0023??

[Dan]

What is the definition of mass percent?

[desertfox23]

Okay so

(.403 moles per kg) x (119.0023 atomic mass of KBr) = 47.9579 grams

1000 grams in a kg.

Would this mean 4.79579% is the mass percent?

I'm confused since it never gave me the amount of H2O

[mjc123]

What is the definition of molality?

[desertfox23]

What is the definition of molality?

All I know is molality = (moles of solute) / (kg of solvent)

It doesn't give me the kg of H2O.

I've made attempts to solve it and I'm not getting any help. I honestly don't know what to do. It may be clear as day to many of you but it's been almost a decade since I took anything related to chemistry and almost 5 years since my last math class.

[mjc123]

That's like the man who was arrested for driving at 100 miles per hour and said "How can you know? I haven't been driving for an hour yet."
You don't need to know how many kg of water you have. You know how many moles of KBr per kg - you have worked that out, presumably correctly, I haven't checked. And you have converted this to g KBr per kg water. It is a short step from here to the mass % KBr. (Hint: it's not 4.79579%. Why not?)

[desertfox23]

That's like the man who was arrested for driving at 100 miles per hour and said "How can you know? I haven't been driving for an hour yet."
You don't need to know how many kg of water you have. You know how many moles of KBr per kg - you have worked that out, presumably correctly, I haven't checked. And you have converted this to g KBr per kg water. It is a short step from here to the mass % KBr. (Hint: it's not 4.79579%. Why not?)

Wish I knew. I wouldn't be asking if I did.

I assume per kg means just using 1000g as a basis. 47.79579g of KBr made me assume 4.79579 would be the percentrage.

[mjc123]

Molality is moles of solute per kg of solvent.
Mass % is g solute per 100g of solution.
So if you have 47.96 g KBr in 1000 g water, what is the mass of the solution? What is the mass of KBr as a percentage of this?

[desertfox23]

Molality is moles of solute per kg of solvent.
Mass % is g solute per 100g of solution.
So if you have 47.96 g KBr in 1000 g water, what is the mass of the solution? What is the mass of KBr as a percentage of this?

So am I dividing 47.96g by 100g and then multiplying it by 10?

or am I dividing 47.96 by 1047.96 and then multiplying it by 100?

[Borek]

So am I dividing 47.96g by 100g and then multiplying it by 10?

or am I dividing 47.96 by 1047.96 and then multiplying it by 100?

You are trying to guess instead of trying to follow a systematic approach. I will multiply that by that and perhaps it will work. That is asking for failure I am afraid.

mjc pointed you in the right direction - what is the mass of the solution, if there are 47.96 g of KBr added to 1000 g of solvent?

[desertfox23]

1047.96g I would assume? So much time spent on one question and seems like I have to beg for help.

[Dan]

1047.96g I would assume? So much time spent on one question and seems like I have to beg for help.

Ok, so now you know the mass of the solution and the mass of the solute. Now look up the definition of mass percent and apply it to your question.