Finding mol of H2O in a solution

[joonkimdds]

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Q: "a 1 liter aqueous solution contains 18g of C6H12O6 and 3.2 g CH3OH. Find the mole of H2O."

How do you find the mole of H2O?

 

[Mstoothlady2012]

I am guessing that when glucose reacts with methanol it produces water and something else. But i dont really know how to write out equation and balance it. But if you can figure that out then it is very easy. Convert g --> mol for whichever is the limiting reagent and then find mol:mol ratio for the water:limiting reactant and multiply by that and you should be able to get the answer.

 

[PreDent2009]

1 Liter of H20 = 1000g of H20 I believe, now you can solve for the number of moles in solution. Solve for the number of moles of the other two compounds and put them all over the total seperately to get each seperate mole fraction.

 

[Mstoothlady2012]

1 Liter of H20 = 1000g of H20 I believe, now you can solve for the number of moles in solution. Solve for the number of moles of the other two compounds and put them all over the total seperately to get each seperate mole fraction.

we are not looking for mole fraction and it doesnt say that we have 1 L of H2O

 

[doc toothache]

Q: "a 1 liter aqueous solution contains 18g of C6H12O6 and 3.2 g CH3OH. Find the mole of H2O."

How do you find the mole of H2O?

Since 21.2 g are made of methanol and C6H12O6, the remainder of 978.8 is H2O.
Then moles H2O= 978.8/18=54.38.

 

[osimsDDS]

Since 21.2 g are made of methanol and C6H12O6, the remainder of 978.8 is H2O.
Then moles H2O= 978.8/18=54.38.

Again you amaze me by using pure logic, u made it extremely easy to understand by taking the easiest route...thanks

 

[joonkimdds]

Since 21.2 g are made of methanol and C6H12O6, the remainder of 978.8 is H2O.
Then moles H2O= 978.8/18=54.38.

That's exactly what I got
"solution mean it's solute + solvent.
So I think we should subtract volume of solutes from 1 L to find the volume of solvent.
18 g of solute is 0.018L, 3.2g of solute is 0.0032L.
1 - 0.018 - 0.0032 = 0.9788 L of water
0.9788 L of water = 978.8g and that's 54.38 mol of water." - said JoonKimDDS 🙂

but!!! I am sorry to disappoint you all. 54.38 isn't the answer.

According to Dr. Romano (DAT destroyer)
"1L = 1000g H2O = 55.6moles of H2O"

This is actually the question number GenChem#68 of DAT destroyer.

I am confused. The question said solution is 1L, it didn't say solvent is 1L.
So how can we assume that solvent(water) has 1L?
why shouldn't we subtract the amount of solutes?
Again, solution means solute + solvent and we want solvent only.

 

[PreDent2009]

Like i had noted earlier, when a question says 1 L of aqueous solution, that assumes 1 L of H20, which is 1000g of water, which I think is roughly 53.3 moles or something like that. I did this question while studying for the DAT and that is how you solve it. In Gen Chem problems, you can assume a lot of things, especially when dealing with H2O. This questions actually directly asks to solve for the mole fraction of one of those 2 compounds if I remember correctly.

 

[joonkimdds]

Like i had noted earlier, when a question says 1 L of aqueous solution, that assumes 1 L of H20, which is 1000g of water, which I think is roughly 53.3 moles or something like that. I did this question while studying for the DAT and that is how you solve it. In Gen Chem problems, you can assume a lot of things, especially when dealing with H2O. This questions actually directly asks to solve for the mole fraction of one of those 2 compounds if I remember correctly.

but that really doesn't make sense though 🙁
are we assuming that test makers don't know enough about general chemistry?
or are we assuming that volume of water is always equal to volume of solution?

 

[doc toothache]

but!!! I am sorry to disappoint you all. 54.38 isn't the answer.

According to Dr. Romano (DAT destroyer)
"1L = 1000g H2O = 55.6moles of H2O"

This is actually the question number GenChem#68 of DAT destroyer.

I am confused. The question said solution is 1L, it didn't say solvent is 1L.
So how can we assume that solvent(water) has 1L?
why shouldn't we subtract the amount of solutes?
Again, solution means solute + solvent and we want solvent only.

You are so confused that it is difficult to tell just exactly what it is that you are trying to say. The figure of 55.6 that you quoted represents the number of moles of water in one liter (~1kg ) of water. When one liter of solution containing 18g of a monosaccharide and 3.2 g of methanol is made, the moles of water decreases to 54.38. Aqueous refers to water. A solvent is any substance that dissolves another to form a solution. Calculating the mole fraction shouldn't be a problem. The numbers were chosen on purpose to simplify calculations. You can divide 18/180 to get moles of glucose and 3.2/32 to get moles of methanol. Your total moles is 54.58. Mole fraction water is 54.38/55.58=0.9963 and that for glucose and methanol would be 0.1/54.58=0.0018. When you figure out the correct answer do let us know because some of us are dying to find out.

 

[joonkimdds]

You are so confused that it is difficult to tell just exactly what it is that you are trying to say. The figure of 55.6 that you quoted represents the number of moles of water in one liter (~1kg ) of water. When one liter of solution containing 18g of a monosaccharide and 3.2 g of methanol is made, the moles of water decreases to 54.38. Aqueous refers to water. A solvent is any substance that dissolves another to form a solution. Calculating the mole fraction shouldn't be a problem. The numbers were chosen on purpose to simplify calculations. You can divide 18/180 to get moles of glucose and 3.2/32 to get moles of methanol. Your total moles is 54.58. Mole fraction water is 54.38/55.58=0.9963 and that for glucose and methanol would be 0.1/54.58=0.0018. When you figure out the correct answer do let us know because some of us are dying to find out.

This question was actually asking for mole fraction of C6H12O6. (Even though I asked to find the mol of water)
To do that we need to find mole of C6H12O6 and divide it by total mol.
I know how to find the mole of C6H12O6 and CH3OH.
But just like you, i found that the number of mole of water is 54.38.
so in our theory, the mole of C6H12O6 should be
0.1 / (54.38+0.1+0.1) = 0.1/ 54.58
**0.1 is mol of C6H12O6 and also it's mol of CH3OH.

But the answer in the book is 0.1/55.8
It says
0.1 / (55.6+0.1+0.1) = 0.1/55.8
It says the reason the mol of water is 55.6 is because there is 1L of water.

so basically, i changed the question a little bit and asked how I find the mol of water
since the mole I found(=what u found) is different from what Dr. Romano found in his book.

 

[doc toothache]

This question was actually asking for mole fraction of C6H12O6. (Even though I asked to find the mol of water)
To do that we need to find mole of C6H12O6 and divide it by total mol.
I know how to find the mole of C6H12O6 and CH3OH.
But just like you, i found that the number of mole of water is 54.38.
so in our theory, the mole of C6H12O6 should be
0.1 / (54.38+0.1+0.1) = 0.1/ 54.58
**0.1 is mol of C6H12O6 and also it's mol of CH3OH.

But the answer in the book is 0.1/55.8
It says
0.1 / (55.6+0.1+0.1) = 0.1/55.8
It says the reason the mol of water is 55.6 is because there is 1L of water.

so basically, i changed the question a little bit and asked how I find the mol of water
since the mole I found(=what u found) is different from what Dr. Romano found in his book.

There is always a great deal of confusion when it comes to making solutions, and, if the question is worded as you posted, even in the case of Dr. Romano. The solutions can be made two different ways. In the first, (which is the way you worded the problem) using a volumetric flask, after placing glucose and methanol, water is added to the 1 liter mark. In the second case the methanol and glucose are added to 1 liter of water, making the final volume greater than 1 liter. Ignoring the effect of specific gravities of glucose and methanol on the density of water, the final volume should be around 1021.2 ml. This is the reason why the answers do not match.

 

[joonkimdds]

There is always a great deal of confusion when it comes to making solutions, and, if the question was is worded as you posted, even in the case of Dr. Romano. The solutions can be made two different ways. In the first, (which is the way you worded the problem) using a volumetric flask, after placing glucose and methanol, water is added to the 1 liter mark. In the second case the methanol and glucose are added to 1 liter of water, making the final volume greater than 1 liter. Ignoring the effect of specific gravities of glucose and methanol on the density of water, the final volume should be around 1021.2 ml. This is the reason why the answers do not match.

Then how should we deal with this type of problem?
I really want to say that the question is not worded correctly since it said solution is 1L, not solvent. But I am not the one who can say what's right or wrong if everyone who studies chemistry follows this rule.

 

[mddang]

1g of a compound does not necessarily equal 1mL of solution... Not everything has the same density.
For example, the density of pure methanol at a certain temperature is 0.7918 g/cm³.
(http://en.wikipedia.org/wiki/Methanol)

For that reason, doc toothache's reasoning isn't completely correct.



However, 1g of water does equal about 1mL

Unless stated otherwise, I would assume the 1L is all water and ignore the volume occupied by the other compounds.

 

[doc toothache]

Then how should we deal with this type of problem?
I really want to say that the question is not worded correctly since it said solution is 1L, not solvent. But I am not the one who can say what's right or wrong if everyone who studies chemistry follows this rule.

There is nothing wrong with the way the problem was worded, and it would not make an iota of difference if the word solvent was used. It is best to go with the problem the way it is worded and then worry about the correct answer. If it will make you feel better, you should call Dr. R. and ask him. But even he is not going to be able to squeeze 0.2 moles of another substance in 1 liter of water and still have a final volume of 1 liter.

1g of a compound does not necessarily equal 1mL of solution... Not everything has the same density.
For example, the density of pure methanol at a certain temperature is 0.7918 g/cm³.
(http://en.wikipedia.org/wiki/Methanol)
For that reason, doc toothache's reasoning isn't completely correct.
However, 1g of water does equal about 1mL
Unless stated otherwise, I would assume the 1L is all water and ignore the volume occupied by the other compounds.

In this particular case the specific gravity has absolutely nothing to do with the problem. The second example of how a solution could be made, specific gravity was mentioned and ignored since the 1021.2 ml of solution was an estimated volume only to point to the fact that the volume would have to be greater than 1 liter if Dr. Romano's explanation is to, well, hold water. By using specific gravities of glucose (1.54) and of methanol the exact volume occupied by the mixture could be calculated (1015.69). Since we are dealing with mole fractions what would be the point?

 

[joonkimdds]

If it will make you feel better, you should call Dr. R. and ask him. But even he is not going to be able to squeeze 0.2 moles of another substance in 1 liter of water and still have a final volume of 1 liter.