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August 12, 2020, 04:28:31 pm

Author Topic: Missprint Chapter 3 Question 6  (Read 1396 times)  Share 

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matty.k

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Missprint Chapter 3 Question 6
« on: February 02, 2010, 07:20:06 pm »
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for a 0.20M solution of potassium sulfate K2SO4, calculate the amount, in mol of:

a. potassium ions, K+

c.oxygen atoms

for a. would i just multiply the molarity by the molar mass of K2SO4 in order to get the mass of K2SO4 then once i get the mass use n=m/Mr for potassium? can someone explain this question for me. thanks in advanced
« Last Edit: February 03, 2010, 07:49:41 pm by matty.k »

the.watchman

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Re: Molarity Question
« Reply #1 on: February 02, 2010, 07:23:13 pm »
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Okay, I think the amount in mol of potassium, is mol
This is because mol is a measure of the number of molecules.

Same for oxygen
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appianway

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Re: Molarity Question
« Reply #2 on: February 02, 2010, 07:32:26 pm »
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Yup, the logic's right, but the number of oxygen atoms should be 0.80 mol, seeing as there are 4 oxygen atoms per sulphate ion.

Martoman

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Re: Molarity Question
« Reply #3 on: February 02, 2010, 07:33:09 pm »
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Keep in mind here there isn't enough information to complete the problem. However, what the.watchman did was assume 1L of K2SO4, as .2M means .2 mols for every 1 litre.
There is obviously a misprint in your textbook.
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the.watchman

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Re: Molarity Question
« Reply #4 on: February 02, 2010, 07:33:33 pm »
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Yup, the logic's right, but the number of oxygen atoms should be 0.80 mol, seeing as there are 4 oxygen atoms per sulphate ion.

LOL, i meant the same PROCESS
never mind... :P
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appianway

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Re: Molarity Question
« Reply #5 on: February 02, 2010, 07:35:20 pm »
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Yeah, it relies on the volume, although you can find [K+] and [SO4 2-].

Yes, I knew you meant the same process which was why I said the logic's right :P

And welcome back. :)

the.watchman

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Re: Molarity Question
« Reply #6 on: February 02, 2010, 07:36:33 pm »
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And welcome back. :)

Thanks, it's good to be back (except the school catch-up work...)
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matty.k

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Re: Molarity Question
« Reply #7 on: February 02, 2010, 07:50:35 pm »
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the back of the book said that the answer to question a. is 0.10mol

the.watchman

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Re: Molarity Question
« Reply #8 on: February 02, 2010, 07:54:28 pm »
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the back of the book said that the answer to question a. is 0.10mol

Ohhh ... any other information?
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matty.k

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Re: Molarity Question
« Reply #9 on: February 02, 2010, 08:00:49 pm »
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ill type the question agian

for a 0.20M solution of potassium sulfate, K2SO4 calculate the amount in mol of

a.potaasium ions, K+

i approached this question by multiplying the molarity by the molar mass of K2SO4 to get the mass of K2SO4, then use n=m/Mr to get the amount of potassium ions but my answer isn't the same as bob's

Martoman

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Re: Molarity Question
« Reply #10 on: February 02, 2010, 08:11:01 pm »
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Q6.
For a 0.20 M solution of potassium sulfate, K2SO4, calculate the number of moles of:
a   potassium ions, K+
b   sulfate ions, SO42
c   oxygen atoms
A6.
a   n(K2SO4) = C V = 0.20 0.250 = 0.050 mol
n(K+) = 2 n(K2SO4) = 2 0.050 = 0.10 mol
b   n(SO42-) = n(K2SO4) = 0.050 mol
c   n(S) = n(K2SO4) = 0.050 mol


Worked solutions from heinneman. Obviously, as i said above, they forgot to print the .25L.
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the.watchman

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Re: Molarity Question
« Reply #11 on: February 02, 2010, 08:12:49 pm »
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Haha, that makes sense now!
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akira88

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Re: Molarity Question
« Reply #12 on: February 02, 2010, 08:54:23 pm »
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Haha I came across this question as well and I was really confused why I was getting the answer wrong... until the teacher said there was a misprint :)
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matty.k

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Re: Molarity Question
« Reply #13 on: February 03, 2010, 03:44:14 pm »
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whoa thanks alot, its all clear to me thanks agian :D