kaplan solutions questions:P

[Hexon]

referrring to kaplan's general chemistry booklet, the first practice question:
an aqueous soln is prepared via mixing 70g of a solid into 100g of water; the soln has a bp of 101.11C. what's the molar mass of the solute?
(kb=0.512C)

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[MShopes]

referrring to kaplan's general chemistry booklet, the first practice question:
an aqueous soln is prepared via mixing 70g of a solid into 100g of water; the soln has a bp of 101.11C. what's the molar mass of the solute?
(kb=0.512C)

Molar mass=Bp of solution*grams of solute/Delta T b*kg of solvent.

Delta T b=I*Molality (moles of solute/kg of solute)*kb

Only thing I would need is the moles of solute or at least the identity of the solid so that I can know what is its molar mass and therefore its moles. And then the problem becaomes easy

 

[MT Headed]

This question cannot be answered with the information given. You would need to know the "I", the van't hoff factor (actual or theoretical) in order to answer the problem. Boiling point elevation is a colligative property, and depends on the moles of particles per mass of solvent. You need to know how many moles of particles you get when one mole of solute dissolves in water. table sugar=1, table salt=2, etc.

Step A: boiling point elevation equation -> molality
deltaT = Kb x molality x i
You know deltaT, you know Kb, and you'll need to get "i" somewhere, then you can solve for the molality in units of (moles solute/kg solvent)

Step B: molality -> moles
molality = (moles solute/kg solvent)
you know molality, you know kg of solvent, solve for moles of solute

Step C: moles -> molar mass
molar mass = (g substance/moles substance)
you know g substance, you know moles of substance, solve for molar mass


As a meta discussion about how to solve these general types of chemistry problems, think about what they want, and what you already know.
(1) they want molar mass, and you already have mass, so once you find moles you are golden
(2) this is a boiling point elevation problem, so dig up the boiling point elevation equation and realize you could probably solve for molality
(3) hope like hell they gave you enough additional information to connect the molality of #2 to the number of moles you need in #1. yup. whew!

 

[MShopes]

This question cannot be answered with the information given. You would need to know the "I", the van't hoff factor (actual or theoretical) in order to answer the problem. Boiling point elevation is a colligative property, and depends on the moles of particles per mass of solvent. You need to know how many moles of particles you get when one mole of solute dissolves in water. table sugar=1, table salt=2, etc.

Step A: boiling point elevation equation -> molality
deltaT = Kb x molality x i
You know deltaT, you know Kb, and you'll need to get "i" somewhere, then you can solve for the molality in units of (moles solute/kg solvent)

Step B: molality -> moles
molality = (moles solute/kg solvent)
you know molality, you know kg of solvent, solve for moles of solute

Step C: moles -> molar mass
molar mass = (g substance/moles substance)
you know g substance, you know moles of substance, solve for molar mass


As a meta discussion about how to solve these general types of chemistry problems, think about what they want, and what you already know.
(1) they want molar mass, and you already have mass, so once you find moles you are golden
(2) this is a boiling point elevation problem, so dig up the boiling point elevation equation and realize you could probably solve for molality
(3) hope like hell they gave you enough additional information to connect the molality of #2 to the number of moles you need in #1. yup. whew!

I agree completely. Only thing you need to know is the I value, from there, you can get the molality and after that you can solve for moles of solute. Ones you have the moles of solute, plug it in the formula I provided above. The identity of solid (molecular formula) is not needed like I mentioned above because fortunately, they provided the Delta T b so you would be golden if you have the I factor. Just one letter is stopping you from solving the problem.

Look for the molecular formula of the compound that you are trying to put in water, maybe they provided it and then the I factor can be known by just looking at the molecular formula.

 

[Hexon]

thanks a lot, guys

 

[Hexon]

hey guys, got another solutions q that's making me tear my hair out; (why the heck doesn't kaplan explain things step by step in the solutions is beyond me)

ok, p172 if u got the manual; q14:
body of water that's polluted with lead ions at 30 ppb, what's the concn of lead in molarity?
density of water=1g/ml.

i got as far as considering 30ppb as being equivalent to 30g of Pb ions in 10^9g of soln, and finding number of moles of Pb:
n=m/MM
n=30g/207g
n=0.145 moles of Pb

but then i got stuck after that,

 

[MT Headed]

Well, molarity is what, (moles of solute/liter of soln)?

I looks like you found the number of moles of solute in an arbitrary 10^9g of solution.

Now go figure out how many liters of water are in this same 10^9g of solution, and divide. There's your answer, in units of (this many moles/this many liters), or (moles/liter).

Good job figuring out that ppb is a (grams of solute/billion grams of solution) ratio by the way. Most people get it wrong at step one. Parts per billion doesn't count "parts", it counts mass.

 

[Hexon]

Well, molarity is what, (moles of solute/liter of soln)?

I looks like you found the number of moles of solute in an arbitrary 10^9g of solution.

Now go figure out how many liters of water are in this same 10^9g of solution, and divide. There's your answer, in units of (this many moles/this many liters), or (moles/liter).

Good job figuring out that ppb is a (grams of solute/billion grams of solution) ratio by the way. Most people get it wrong at step one. Parts per billion doesn't count "parts", it counts mass.

yep, got the answer as a followup:

so we got as far as considering 30ppb as being equivalent to 30g of Pb ions in 10^9g of soln, and finding number of moles of Pb:
n=m/MM
n=30g/207g
n=0.145 moles of Pb

finding molarity:

molarity=moles of solute/litre of solution
considering 10^9g=10^6 litres since for water 1kg=1 litre.

molarity=0.145moles/10^6litres=1.45 x 10^-7