starsbeneathme

2+ Year Member
Jan 5, 2016
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Pre-Dental
the questions says

at 0°C gas is dissolved in H2O at a mol fraction of 2 x 10^-18. What is the molality of the solution given the molecular weight of the gas is 32 g/mol.

so we know:
molality = mols of solute/kg of solvent.

so the explanation in the book is that you can assume that there is 1 mol of water, which weights 0.018 kg.
the explanation goes on to say that to find molality, it would m = (2 x 10^-18)/ 0.018 = 1 x 10^-16

i'm confused about why we can assume that there is 2 x 10^-18 of the solute when we also have to consider the mols of the water since mol fraction = mols of solute/ total mols

..so, technically, shouldn't the formula for mol fraction be... mols solute/ (mols of solute + 1 molH2O) = 2 x 10^-18
 

Mrhyde

Becoming Dr. Jekyll
Apr 13, 2015
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This is my attempt at understanding this (maybe I am wrong but i think i am right lol ) :

First the title say organic chem and this is a Gen chem question :)

--- Next you wrote the Mole Fraction formula WRONG. A mole fraction is not just : moles solution/ total moles ? That is where you got confused.

The formula for a mole fraction is : Moles Solute / total Moles Solution ....................... You cant do mols solute/ (mols of solute + 1 molH2O) Because a mole fraction like I just wrote is just supposed to be over moles of SOLUTION, and you put it as solute + solution so its wrong.

Now to get to Molality....

Solute is what gets dissolved so that is the gas which is in a mole fraction of : 2 x 10^-18 moles solute / 1 mole solution

We have the numerator correct now we need to figure out how to get moles of solution into kg solvent somehow?

2 x 10^-18 moles * 32 grams/ 1 mole = 6.4*10^-17 grams solute

..................................And we know that: grams solute - grams solution = grams solvent

So put it in kg
6.4*10^-17 grams solute * 1 kg/10^3 grams = 6.4*10^-20 kg solute



So now lets get grams of solution:

H20
1 mole H20 * 18 grams/ 1 mole = 18 grams H20

18 grams H20 * 1 kg/10^3 grams = .018 Kg solution

..................................kg solute - kg solution = kg solvent

But when we subtract

6.4*10^-20 grams solute - .018 Kg solution = the number is so small that it remains as .018 Kg

--> so now we have the Kg solvent we needed which is .018 kg solvent

------> (2 x 10^-18) moles solute/ 0.018 kg solvent = 1 x 10^-16 m
 
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ChewyDrop

2+ Year Member
Jun 30, 2015
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This is my attempt at understanding this (maybe I am wrong but i think i am right lol ) :

First the title say organic chem and this is a Gen chem question :)

--- Next you wrote the Mole Fraction formula WRONG. A mole fraction is not just : moles solution/ total moles ? That is where you got confused.

The formula for a mole fraction is : Moles Solute / total Moles Solution ....................... You cant do mols solute/ (mols of solute + 1 molH2O) Because a mole fraction like I just wrote is just supposed to be over moles of SOLUTION, and you put it as solute + solution so its wrong.

Now to get to Molality....

Solute is what gets dissolved so that is the gas which is in a mole fraction of : 2 x 10^-18 moles solute / 1 mole solution

We have the numerator correct now we need to figure out how to get moles of solution into kg solvent somehow?

2 x 10^-18 moles * 32 grams/ 1 mole = 6.4*10^-17 grams solute

..................................And we know that: grams solute - grams solution = grams solvent

So put it in kg
6.4*10^-17 grams solute * 1 kg/10^3 grams = 6.4*10^-20 kg solute



So now lets get grams of solution:

H20
1 mole H20 * 18 grams/ 1 mole = 18 grams H20

18 grams H20 * 1 kg/10^3 grams = .018 Kg solution

..................................kg solute - kg solution = kg solvent

But when we subtract

6.4*10^-20 grams solute - .018 Kg solution = the number is so small that it remains as .018 Kg

--> so now we have the Kg solvent we needed which is .018 kg solvent

------> (2 x 10^-18) moles solute/ 0.018 kg solvent = 1 x 10^-16 m
Very well explained! My exact thought process!
 
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OP
starsbeneathme

starsbeneathme

2+ Year Member
Jan 5, 2016
53
13
Status
Pre-Dental
This is my attempt at understanding this (maybe I am wrong but i think i am right lol ) :

First the title say organic chem and this is a Gen chem question :)

--- Next you wrote the Mole Fraction formula WRONG. A mole fraction is not just : moles solution/ total moles ? That is where you got confused.

The formula for a mole fraction is : Moles Solute / total Moles Solution ....................... You cant do mols solute/ (mols of solute + 1 molH2O) Because a mole fraction like I just wrote is just supposed to be over moles of SOLUTION, and you put it as solute + solution so its wrong.

Now to get to Molality....

Solute is what gets dissolved so that is the gas which is in a mole fraction of : 2 x 10^-18 moles solute / 1 mole solution

We have the numerator correct now we need to figure out how to get moles of solution into kg solvent somehow?

2 x 10^-18 moles * 32 grams/ 1 mole = 6.4*10^-17 grams solute

..................................And we know that: grams solute - grams solution = grams solvent

So put it in kg
6.4*10^-17 grams solute * 1 kg/10^3 grams = 6.4*10^-20 kg solute



So now lets get grams of solution:

H20
1 mole H20 * 18 grams/ 1 mole = 18 grams H20

18 grams H20 * 1 kg/10^3 grams = .018 Kg solution

..................................kg solute - kg solution = kg solvent

But when we subtract

6.4*10^-20 grams solute - .018 Kg solution = the number is so small that it remains as .018 Kg

--> so now we have the Kg solvent we needed which is .018 kg solvent

------> (2 x 10^-18) moles solute/ 0.018 kg solvent = 1 x 10^-16 m

thank you for the explanation-- your answer is correct :)

i don't think i'm wrong about the mol fraction though..it is mol of solute/ mols of solute + mols of solvent because 1 mol solution = mols of solute + mols of solvent.
but since for these problems, you generally assume you have one mol of solution, and since the mols of the solute is so small (2 x 10^-18), you can assume there's 1 mol of H2O, the solvent.