F325 Jan 13 Question 8 help??

Here's the question:

(c) A student carries out an investigation into the oxidation states of vanadium as outlined below.
Stage 1
Stage 2
Stage 3
A 0.126g sample of vanadium metal is completely reacted with acid to form a yellow solution. The solution is made up to 50.0 cm3 in a volumetric flask.
This yellow solution contains VO – ions with vanadium in the +5 oxidation state.
3
The yellow solution is reduced to form a violet solution containing Vn+ ions.
This 50.0 cm3 violet solution contains vanadium in the +n oxidation state.
10.0 cm3 of the violet solution is titrated with 2.25 × 10–2 mol dm–3 KMnO4(aq).
13.2 cm3 of KMnO4(aq) are required to reach the end-point.
In the titration,
• Vn+ ions are oxidised back to VO – ions. 3
• MnO – ions are reduced: 4
MnO –(aq) + 8H+(aq) + 5e– 42
Mn2+(aq) + 4H O(l)

(ii) Analyse the student’s results as follows:
• Determine the value of n in the Vn+ ions formed in Stage 2
• Construct an equation for the reaction that takes place during the titration. Show all your working.

In the mark scheme it says to find the moles of Vandium I need to divide 0.126/50.9, but I have no idea where the value 50.9 came from?? I feel like Im missing something really obvious haha. Any ideas?

Reply 1

Kool_Panda

12

Original post by Chloe-

Here's the question:

(c) A student carries out an investigation into the oxidation states of vanadium as outlined below.
Stage 1
Stage 2
Stage 3
A 0.126g sample of vanadium metal is completely reacted with acid to form a yellow solution. The solution is made up to 50.0 cm3 in a volumetric flask.
This yellow solution contains VO – ions with vanadium in the +5 oxidation state.
3
The yellow solution is reduced to form a violet solution containing Vn+ ions.
This 50.0 cm3 violet solution contains vanadium in the +n oxidation state.
10.0 cm3 of the violet solution is titrated with 2.25 × 10–2 mol dm–3 KMnO4(aq).
13.2 cm3 of KMnO4(aq) are required to reach the end-point.
In the titration,
• Vn+ ions are oxidised back to VO – ions. 3
• MnO – ions are reduced: 4
MnO –(aq) + 8H+(aq) + 5e– 42
Mn2+(aq) + 4H O(l)

(ii) Analyse the student’s results as follows:
• Determine the value of n in the Vn+ ions formed in Stage 2
• Construct an equation for the reaction that takes place during the titration. Show all your working.

In the mark scheme it says to find the moles of Vandium I need to divide 0.126/50.9, but I have no idea where the value 50.9 came from?? I feel like Im missing something really obvious haha. Any ideas?

Mr of Vanadium?

Reply 2

username1171861

So read the first line of the question, and see what you've got:

0.126g of Vanadium Metal - So just plain Vanadium.

Now think what they could do with this number.

I'll tell you if you're still stuck but think about what you can do with the mass.