Question

Volume strength of 1 M solution of \[{{H}_{2}}{{O}_{2}}\] is:
A. 11.2
B. 22.4
C. 10.8
D. 21.6

Answer 1

Hint: Volume strength is concentration of \[{{H}_{2}}{{O}_{2}}\] in terms of volumes of oxygen gas based on its decomposition to form water and oxygen.

Complete step by step answer:

So, for finding volume strength we have to write a balanced reaction of \[{{H}_{2}}{{O}_{2}}\] decomposition to form water and oxygen.
So, here is the balanced reaction of \[{{H}_{2}}{{O}_{2}}\] decomposition to form water and oxygen:\[2{{H}_{2}}{{O}_{2}}\to 2{{H}_{2}}O+{{O}_{2}}\]
In \[{{H}_{2}}{{O}_{2}}\] oxidation state of oxygen is “-1”, because if we apply the rule that sum of oxidation number of all the atoms in a neutral compound is zero and here oxidation state of hydrogen atom is +1, so oxygen atom has -1 oxidation state. Similarly in \[{{H}_{2}}O\] oxidation state of oxygen is “-2”, and in \[{{O}_{2}}\] oxidation state is “0”.
So, from a balanced reaction of \[{{H}_{2}}{{O}_{2}}\] we can see that 2 moles of \[{{H}_{2}}{{O}_{2}}\] give 1 mole of \[{{O}_{2}}\].
Therefore, 1 mole of \[{{H}_{2}}{{O}_{2}}\] will give 11.2 L of \[{{O}_{2}}\] gas, because we know that 1 mole of oxygen at STP has a volume of 22.4 L.
So, Volume strength of \[{{H}_{2}}{{O}_{2}}\] = molarity (molarity is defined as number of moles of solute per litre of volume) x volume of oxygen from 1 mole of \[{{H}_{2}}{{O}_{2}}\] = 1 x 11.2 =11.2 volume
Answer is “B”.
“A” is incorrect because 22.4 L is for 2moles of \[{{H}_{2}}{{O}_{2}}\].

Note: Don’t make mistakes in balancing the \[{{H}_{2}}{{O}_{2}}\] decomposition reaction. Carefully determine the oxidation state of oxygen.