Anon-stoichiometric compound is a compound that cannot be represented by a small whole-number ratio of atoms, usually because of point defects in the crystal lattice. what is the average oxidation state of vanadium in ? what is the average state of vanadium in vo 1.19? if each vanadium atom has either a 2 or 3 oxidation state in this compound, what percentage of the vanadium atoms are in the lower oxidation state?

Average oxidation state = VO1.19
Oxygen is-2. Then 1.19 (-2) = -2.38
Average oxidation state of V is +2.38 

Consider 100 formula units of VO1.19
There would be 119 Oxide ions = Each oxide is -2. Total charge = -2(119) = -238 
The total charge of all the vanadium ions would be +238. 
Let x = number of of V+2 
Then 100 – x = number of V+3 
X(+2) + 100-x(+3) = +238 
2x + 300 – 3x = 238 
-x = 238-300 = -62 
x = 62
 
Thus 62/100 are V+2 
62/100 * 100 = 62%

62 % is the percentage of the vanadium atoms are in the lower oxidation state. Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help. 

Answer is to avoid experimental error

True, the standard unit for volume is liters which can be represented by either a lower case "l" or a capital "l"