If the components of air are${{N}_{2}}$ $78%$, ${{O}_{2}}$ $21%$, $Ar$ $0\cdot 9%$ and $C{{O}_{2}}$ $0\cdot 1%$ by volume, the molar mass of air would be (nearest integer):
Answer
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Hint: To find the molar mass of air, we can multiply the given percentage of constituent elements with their respective molar masses. Molar mass is the mass of all the atoms of a molecule in grams per mole.
Complete step by step answer:
For ${{N}_{2}}$, we have molar mass = 28 g/mole where we have calculated it as follows: For one single atom of nitrogen we have atomic no. 7 which is the no. of protons and mass no. is 14 which is the no. of protons and the no. of neutrons.
So, for ${{N}_{2}}$ we will simply multiply the atomic mass no. of single atom by 2 i.e. $14\times 2=28$ g/mole
Similarly for ${{O}_{2}}$: For one single oxygen atom we have atomic mass no. =16, the molar mass for ${{O}_{2}}=16\times 2=32$ g/mol.
Molar Mass for Argon $(Ar)$$=40$g/mol.
For $C{{O}_{2}}$ we have one carbon and two oxygen atoms, the mass no. for Carbon $=12$ and mass no. for Oxygen $=16$, so molar mass of ${{O}_{2}}$$=16\times 2=32$g/mole and molar mass of $C{{O}_{2}}$$=12+32=44$g/mol.
So, the molar mass of air$=$ component $%$ of${{N}_{2}}\times $$\times $ molar mass of ${{N}_{2}}+$ component $%$ of ${{O}_{2}}\times $ molar mass of ${{O}_{2}}+$ component $%$ of $Ar\times $ molar mass of $Ar+$ component $%$ of $C{{O}_{2}}\times $molar mass of$C{{O}_{2}}$
$\Rightarrow \dfrac{78\times 28+21\times 32+0\cdot 9\times 40+0\cdot 1\times 44}{100}$
$\Rightarrow \dfrac{2184+672+36+4\cdot 4}{100}$
$\Rightarrow \dfrac{2896\cdot 4}{100}$
$\therefore 28\cdot 964$
So, the nearest integer value will be 29.
If we have a digit greater than $5$ after the decimal then we go for the next integer value. Here, in the above question, we have $28\cdot 964$ i.e. 9 after the decimal to nearest integer value will be 29 i.e. next to 28.
Note:
Molar mass is given by the mass of the chemical element/compound divided by the amount of substance. So, molar mass of a compound can be calculated by simply adding the standard atomic masses of the constituent atoms.
Atomic masses are given in amu(atomic mass unit) because each proton and each neutron weigh one amu.
Complete step by step answer:
For ${{N}_{2}}$, we have molar mass = 28 g/mole where we have calculated it as follows: For one single atom of nitrogen we have atomic no. 7 which is the no. of protons and mass no. is 14 which is the no. of protons and the no. of neutrons.
So, for ${{N}_{2}}$ we will simply multiply the atomic mass no. of single atom by 2 i.e. $14\times 2=28$ g/mole
Similarly for ${{O}_{2}}$: For one single oxygen atom we have atomic mass no. =16, the molar mass for ${{O}_{2}}=16\times 2=32$ g/mol.
Molar Mass for Argon $(Ar)$$=40$g/mol.
For $C{{O}_{2}}$ we have one carbon and two oxygen atoms, the mass no. for Carbon $=12$ and mass no. for Oxygen $=16$, so molar mass of ${{O}_{2}}$$=16\times 2=32$g/mole and molar mass of $C{{O}_{2}}$$=12+32=44$g/mol.
So, the molar mass of air$=$ component $%$ of${{N}_{2}}\times $$\times $ molar mass of ${{N}_{2}}+$ component $%$ of ${{O}_{2}}\times $ molar mass of ${{O}_{2}}+$ component $%$ of $Ar\times $ molar mass of $Ar+$ component $%$ of $C{{O}_{2}}\times $molar mass of$C{{O}_{2}}$
$\Rightarrow \dfrac{78\times 28+21\times 32+0\cdot 9\times 40+0\cdot 1\times 44}{100}$
$\Rightarrow \dfrac{2184+672+36+4\cdot 4}{100}$
$\Rightarrow \dfrac{2896\cdot 4}{100}$
$\therefore 28\cdot 964$
So, the nearest integer value will be 29.
If we have a digit greater than $5$ after the decimal then we go for the next integer value. Here, in the above question, we have $28\cdot 964$ i.e. 9 after the decimal to nearest integer value will be 29 i.e. next to 28.
Note:
Molar mass is given by the mass of the chemical element/compound divided by the amount of substance. So, molar mass of a compound can be calculated by simply adding the standard atomic masses of the constituent atoms.
Atomic masses are given in amu(atomic mass unit) because each proton and each neutron weigh one amu.
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