Question
Question: When 20 ml of 2.5 M \({{\text{H}}_2}{{\text{O}}_2}\)aqueous solutions is decomposed at S.T.P. volume...
When 20 ml of 2.5 M H2O2aqueous solutions is decomposed at S.T.P. volume of oxygen obtained is
A) 56 ml
B) 560 ml
C) 11.2 ml
D) 2.24 vol
Solution
The question mentions two conditions of molarity(i.e., molarity = 2.5 M, and volume = 20 ml) hence we have to find out the third term and see if we can use it to reach our answer.
Formula used: Morality = Volume(L) Number of moles
Complete answer:
First, let’s write the balanced chemical reaction and see what happens when H2O2 is decomposed,
2H2O2→2H2O + O2
From the above reaction, we can conclude that in general, 2 moles H2O2 yield 1 mole of O2
Therefore, 1 mole H2O2 will yield 21 moles of O2
But, to find out how many moles of H2O2 is formed in the given conditions of our question, we will use the formula for molarity to calculate the same,
Morality = Volume(L) Number of moles
⇒Number of moles = Molarity×Volume(L)
We are given two values i.e., molarity and volume but the given volume is in ml we have to convert it to liter (L) by dividing it by 1000 (since 1ml = 10001L). Therefore 20ml = 10001 × 20 L.
Now putting up all the values in the mentioned formula, we will calculate the number of moles of H2O2
⇒Number of moles = 2.5×10001×20
⇒Number of moles = 0.05
Therefore, for the given conditions in our question 0.05 moles of H2O2 will form,
Now we will find out how many moles of O2 will be yielded by 0.05 moles of H2O2
⇒Number of moles of O2 = 21×0.05
= 0.025 moles of O2
We know that 1 mole of any atom has = 22.4 L of volumes.Therefore, 0.025 moles of O2 has volume
⇒ 22.4×0.025 = 0.056 L
As most of the options are given in ml we will have to convert 0.56 L in ml to check if it matches with any given answer,
⇒ 0.056 L = 0.056×1000 =560 ml
Hence the correct answer is option (b)i.e., 560 ml
Note: Carefully read the S.I units of the measured quantity and change them as per the requirements of the question. Likewise, in this question, S.I units were changed two times first ml to L, and the L to ml both in different situations.