
At a given condition, \[1{{ }}mol{{ }}{O_2}\] occupies \[30{{ }}L\] . Volume occupied by $1$ equivalent of \[{O_2}\] is:
A. \[7.5{{ }}L\]
B. \[15{{ }}L\]
C. \[30.0{{ }}L\]
D. \[22.4{{ }}L\]
Answer
554.7k+ views
Hint: The atomic weight of oxygen is \[16\] and we know that it has the valence of two, so using this we will first calculate the number of grams in one equivalent of \[{O_2}\] . Then using the unitary method we will calculate the answer to our question.
Formula used:
$E = \dfrac{{at.wt}}{n}$
Where, $E$ is Equivalent weight
$at.wt$ is atomic weight
$n$ is valence
Complete answer:
Firstly let us calculate the equivalent weight which is defined as the atomic weight of a substance divided by its valence or the combining power. Hence one equivalent of \[{O_2}\] would be:
$1E({O_2}) = \dfrac{{at.wt({O_2})}}{{n({O_2})}}$
$ \Rightarrow 1E({O_2}) = \dfrac{{16}}{2} = 8g$
Now we know that mass of one mole of \[{O_2}\] is \[32{{ }}grams\]
At the given conditions we have
\[1{{ }}mol{{ }}{O_2}\] occupies \[ = 30{{ }}L\] of the volume
Therefore, \[32{{ }}gram\] of \[{O_2}\] will occupy \[ = {{ }}30{{ }}L\] of the volume at the given conditions.
Hence, using unitary method
\[1{{ }}gram\] of \[{O_2}\] will occupy \[ = {{ }}\dfrac{{30}}{{32}}L\] of the volume at the given conditions
Now, we have to calculate for one equivalent of \[{O_2}\] , as we have already calculated one equivalent of \[{O_2}\] is equal to \[8{{ }}grams\] ,
Therefore, 8 grams of \[{O_2}\] will occupy \[ = \dfrac{{30}}{{32}}{{ }} \times {{ }}8\]
On calculating the answer would come out to be \[7.5{{ }}L\] which is Option A.
Additional information: It is important to note here that at standard temperature and pressure conditions the volume occupied by one mole of oxygen gas is \[22.4{{ }}litres\] . However here in the question the condition given was not specified and was said to be some ‘given condition’, thus we used the volume value given in the question and there wasn’t a use of the STP volume.
Note:
In reaction stoichiometry, one equivalent is the amount of a substance that reacts with (or is either equivalent to) an arbitrary amount of another substance. It is important to know the valency of the atom/molecule to know its equivalent weight.
Formula used:
$E = \dfrac{{at.wt}}{n}$
Where, $E$ is Equivalent weight
$at.wt$ is atomic weight
$n$ is valence
Complete answer:
Firstly let us calculate the equivalent weight which is defined as the atomic weight of a substance divided by its valence or the combining power. Hence one equivalent of \[{O_2}\] would be:
$1E({O_2}) = \dfrac{{at.wt({O_2})}}{{n({O_2})}}$
$ \Rightarrow 1E({O_2}) = \dfrac{{16}}{2} = 8g$
Now we know that mass of one mole of \[{O_2}\] is \[32{{ }}grams\]
At the given conditions we have
\[1{{ }}mol{{ }}{O_2}\] occupies \[ = 30{{ }}L\] of the volume
Therefore, \[32{{ }}gram\] of \[{O_2}\] will occupy \[ = {{ }}30{{ }}L\] of the volume at the given conditions.
Hence, using unitary method
\[1{{ }}gram\] of \[{O_2}\] will occupy \[ = {{ }}\dfrac{{30}}{{32}}L\] of the volume at the given conditions
Now, we have to calculate for one equivalent of \[{O_2}\] , as we have already calculated one equivalent of \[{O_2}\] is equal to \[8{{ }}grams\] ,
Therefore, 8 grams of \[{O_2}\] will occupy \[ = \dfrac{{30}}{{32}}{{ }} \times {{ }}8\]
On calculating the answer would come out to be \[7.5{{ }}L\] which is Option A.
Additional information: It is important to note here that at standard temperature and pressure conditions the volume occupied by one mole of oxygen gas is \[22.4{{ }}litres\] . However here in the question the condition given was not specified and was said to be some ‘given condition’, thus we used the volume value given in the question and there wasn’t a use of the STP volume.
Note:
In reaction stoichiometry, one equivalent is the amount of a substance that reacts with (or is either equivalent to) an arbitrary amount of another substance. It is important to know the valency of the atom/molecule to know its equivalent weight.
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