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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: matty.k on February 02, 2010, 07:20:06 pm

Title: Missprint Chapter 3 Question 6
Post by: matty.k on February 02, 2010, 07:20:06 pm: 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
Title: Re: Molarity Question
Post by: the.watchman on February 02, 2010, 07:23:13 pm: Okay, I think the amount in mol of potassium, is $2\times0.2$ mol
This is because mol is a measure of the number of molecules.

Same for oxygen
Title: Re: Molarity Question
Post by: appianway on February 02, 2010, 07:32:26 pm: 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.
Title: Re: Molarity Question
Post by: Martoman on February 02, 2010, 07:33:09 pm: 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.
Title: Re: Molarity Question
Post by: the.watchman on February 02, 2010, 07:33:33 pm: Quote from: appianway on February 02, 2010, 07:32:26 pm
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
Title: Re: Molarity Question
Post by: appianway on February 02, 2010, 07:35:20 pm: 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. :)
Title: Re: Molarity Question
Post by: the.watchman on February 02, 2010, 07:36:33 pm: Quote from: appianway on February 02, 2010, 07:35:20 pm
And welcome back. :)

Thanks, it's good to be back (except the school catch-up work...)
Title: Re: Molarity Question
Post by: matty.k on February 02, 2010, 07:50:35 pm: the back of the book said that the answer to question a. is 0.10mol
Title: Re: Molarity Question
Post by: the.watchman on February 02, 2010, 07:54:28 pm: Quote from: matty.k on February 02, 2010, 07:50:35 pm
the back of the book said that the answer to question a. is 0.10mol

Ohhh ... any other information?
Title: Re: Molarity Question
Post by: matty.k on February 02, 2010, 08:00:49 pm: 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
Title: Re: Molarity Question
Post by: Martoman on February 02, 2010, 08:11:01 pm: 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.
Title: Re: Molarity Question
Post by: the.watchman on February 02, 2010, 08:12:49 pm: Haha, that makes sense now!
Title: Re: Molarity Question
Post by: akira88 on February 02, 2010, 08:54:23 pm: 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 :)
Title: Re: Molarity Question
Post by: matty.k on February 03, 2010, 03:44:14 pm: whoa thanks alot, its all clear to me thanks agian :D