
The number of mole of oxygen in one litre of air containing $21\% $ oxygen by volume, under standard condition is :
(A). $0.0093mole$
(B). $2.10mole$
(C). $0.186mole$
(D). $0.21mole$
Answer
557.1k+ views
Hint:
$21\% $ of oxygen is present in air by volume means that $21{\text{ ml}}$ of oxygen is present in $100{\text{ ml}}$ volume of air. At STP conditions, $1$ mole contains $22.4{\text{ L}}$of a substance/air.
Complete step by step answer:
As we know that a mole is defined as the amount of substance which has mass equal to gram atomic or molecular mass or if it contains Avogadro no.$\left( {6.022 \times {{10}^{23}}} \right)$ of atoms/molecules.
The atomic mass of an element is the number of times an atom of the element is heavier than an atom of carbon taken as $12$ . The atomic mass of an element expressed in gram is called gram atomic mass and the molecular mass of a substance expressed in gram is called gram molecular mass.
For a gas, a mole is defined as the amount of gas which has a volume of $22.4{\text{ L}}$ or $22400{\text{ ml }}$ at STP.
Now according to question, an air of volume $1{\text{ L}}$ contains $21\% $ of oxygen by volume. This means that amount of oxygen present in air is $21\% $ of $1{\text{ L }}\operatorname{of} {\text{ }}\operatorname{air} {\text{ = }}\dfrac{{21}}{{100}} \times 1$
$ = 0.21L$
This implies that $0.21L$ of oxygen is present in $1{\text{ L}}$ of air. At STP conditions, $1$ mole gas is equal to $22.4{\text{ L}}$ volume.
That is,
$22.4{\text{ L}}$ of oxygen contains mole ${\text{ = 1 mole}}$
$\therefore 0.21{\text{ L}}$ of oxygen mole $ {\text{ = }}\dfrac{{0.21}}{{22.4}} = 0.093{\text{ mole}}$
Thus the required number of moles of oxygen is $0.093mole$.
Hence, option (A) is the correct answer.
Note:The STP Condition is known as standard temperature-pressure condition. The value of temperature at STP is $273K$ and pressure at STP is $1\operatorname{atm} $ . If the pressure is changed to $22.7{\text{ L}}$ that condition is generally known as NTP condition (normal temperature-pressure condition).
$21\% $ of oxygen is present in air by volume means that $21{\text{ ml}}$ of oxygen is present in $100{\text{ ml}}$ volume of air. At STP conditions, $1$ mole contains $22.4{\text{ L}}$of a substance/air.
Complete step by step answer:
As we know that a mole is defined as the amount of substance which has mass equal to gram atomic or molecular mass or if it contains Avogadro no.$\left( {6.022 \times {{10}^{23}}} \right)$ of atoms/molecules.
The atomic mass of an element is the number of times an atom of the element is heavier than an atom of carbon taken as $12$ . The atomic mass of an element expressed in gram is called gram atomic mass and the molecular mass of a substance expressed in gram is called gram molecular mass.
For a gas, a mole is defined as the amount of gas which has a volume of $22.4{\text{ L}}$ or $22400{\text{ ml }}$ at STP.
Now according to question, an air of volume $1{\text{ L}}$ contains $21\% $ of oxygen by volume. This means that amount of oxygen present in air is $21\% $ of $1{\text{ L }}\operatorname{of} {\text{ }}\operatorname{air} {\text{ = }}\dfrac{{21}}{{100}} \times 1$
$ = 0.21L$
This implies that $0.21L$ of oxygen is present in $1{\text{ L}}$ of air. At STP conditions, $1$ mole gas is equal to $22.4{\text{ L}}$ volume.
That is,
$22.4{\text{ L}}$ of oxygen contains mole ${\text{ = 1 mole}}$
$\therefore 0.21{\text{ L}}$ of oxygen mole $ {\text{ = }}\dfrac{{0.21}}{{22.4}} = 0.093{\text{ mole}}$
Thus the required number of moles of oxygen is $0.093mole$.
Hence, option (A) is the correct answer.
Note:The STP Condition is known as standard temperature-pressure condition. The value of temperature at STP is $273K$ and pressure at STP is $1\operatorname{atm} $ . If the pressure is changed to $22.7{\text{ L}}$ that condition is generally known as NTP condition (normal temperature-pressure condition).
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