Missprint Chapter 3 Question 6

[matty.k]

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

[the.watchman]

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

[appianway]

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]

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.

[the.watchman]

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...

[appianway]

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

And welcome back.

[the.watchman]

And welcome back.

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

[matty.k]

the back of the book said that the answer to question a. is 0.10mol

[the.watchman]

the back of the book said that the answer to question a. is 0.10mol

Ohhh ... any other information?

[matty.k]

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]

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.

[the.watchman]

Haha, that makes sense now!

[akira88]

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

[matty.k]

whoa thanks alot, its all clear to me thanks agian