Destroyer Chem #192 please help

[engr2dent]

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I have a question regarding the solution to Destroyer 2010 #182 on chemistry.

182) At 0*C a gas is dissolved in H20 at a mole fraction of 2x10^-18. What is the molality of this solution given the molecular weight of the gas is 32 g/mole.

The solution is:
Assume 1 mole of H20, thus 1 mole H2O x 18g mole/1 mole = 18g = .018 kg. Then molality = (2x10^-18) / (0.018 kg) = 1 x 10^-16.

I am confused as to how a person should know to assume 1 mole of H20. Any insight to this? Thanks!

 

[jirotrom]

I havent reviewed my chemistry yet but let me take a stretch... anyone please correct me if im wrong. I'm assuming this is at STP since it says 0 degrees celcius. So we are following ideal gas laws and they are using a mole fraction... which is why I believe you can assume 1 mole for H2O... just a guess though as I could be totally wrong... lol.

In about 3 weeks my chem knowledge will be back 😀

 

[UCB05]

The assumption is made because 1mol is an easy number to work with. The actual number is inconsequential, as any more or less water will increase or decrease the molar amount of gas by the same coefficient and lead to the exact same answer.

assume 1 mol H2O
Your setup would be (1)(2x10^-18) / [(1)(0.018 kg)]

assume 2 mols H2O.
Your setup would be (2)(2x10^-18)/ [(2)(0.018)]

assume 0.5 mols H2O
Your setup would be (0.5)((2x10^-18) / [(0.5)(0.018 kg)]

 

[Cool Beans]

I have a question regarding the solution to Destroyer 2010 #182 on chemistry.

182) At 0*C a gas is dissolved in H20 at a mole fraction of 2x10^-18. What is the molality of this solution given the molecular weight of the gas is 32 g/mole.

The solution is:
Assume 1 mole of H20, thus 1 mole H2O x 18g mole/1 mole = 18g = .018 kg. Then molality = (2x10^-18) / (0.018 kg) = 1 x 10^-16.

I am confused as to how a person should know to assume 1 mole of H20. Any insight to this? Thanks!

A mole fraction is an intensive property and is therefore independent of sample size so you could have picked any sample size of your liking (same is true in percent composition). Personally I chose 1kg of water in solving this one.

1kg H2O ~ 55moles H2O because 1000g/(18g/mol)

mol x/(mol H2O+mol x)=2x10-18 definition of mol fraction
The mol of x are negligible compared to water so we'll ignore it in the denominator leading to:

mol x/55=2x10-18 and so
molx = (55)x(2x10-18)
molx ~ 100x10-18=1x10-16

molality = (mol solute)/(kg solvent)
molality = (1x10-16)/1 kg = 1x10-16 molal

I solved it with a different sample size than the Destroyer answer key but again the key is that you can choose any sample size you want.

 

[engr2dent]

The assumption is made because 1mol is an easy number to work with. The actual number is inconsequential, as any more or less water will increase or decrease the molar amount of gas by the same coefficient and lead to the exact same answer.

assume 1 mol H2O
Your setup would be (1)(2x10^-18) / [(1)(0.018 kg)]

assume 2 mols H2O.
Your setup would be (2)(2x10^-18)/ [(2)(0.018)]

assume 0.5 mols H2O
Your setup would be (0.5)((2x10^-18) / [(0.5)(0.018 kg)]

A mole fraction is an intensive property and is therefore independent of sample size so you could have picked any sample size of your liking (same is true in percent composition). Personally I chose 1kg of water in solving this one.

1kg H2O ~ 55moles H2O because 1000g/(18g/mol)

mol x/(mol H2O+mol x)=2x10-18 definition of mol fraction
The mol of x are negligible compared to water so we'll ignore it in the denominator leading to:

mol x/55=2x10-18 and so
molx = (55)x(2x10-18)
molx ~ 100x10-18=1x10-16

molality = (mol solute)/(kg solvent)
molality = (1x10-16)/1 kg = 1x10-16 molal

I solved it with a different sample size than the Destroyer answer key but again the key is that you can choose any sample size you want.

Thank you. This makes perfect sense.

 

[luv8724]

Then what happen to 32g/mol?

 

[UCB05]

Since molality is mols solute/kg solvent and you're already given the mol fraction, the molar mass is just a bit of extraneous info you don't need.

 

[engr2dent]

Since molality is mols solute/kg solvent and you're already given the mol fraction, the molar mass is just a bit of extraneous info you don't need.

Yes, that's what the book said.

 

[predent7]

and.. for mole fraction, use 1mole as a basis.