
The number of atoms in $10litres$ of ammonia are: (Molecular weight of $N = 14g/mole,H = 1g/mol$ ).
$A.$ $0.0893 \times 6.023 \times {10^{23}}atoms$
\[B.\] \[0.786 \times 6.023 \times {10^{23}}atoms\]
$C.$ $3.572 \times 6.023 \times {10^{23}}atoms$
$D.$ $1.786 \times 6.023 \times {10^{23}}atoms$
Answer
554.1k+ views
Hint: The given question can be solved by considering, the volume of one mole of any gas is known as molar volume is equal to $22.4L$ at standard temperature and pressure. Molar volume allows conversion to be made between volume and moles of any gases at standard temperature.
Complete answer:
We know any type of gas occupies approximately $22.4litres$ of space at standard temperature pressure, we also know $1mole$ of contain Avogadro’s number of molecules.
So, we can say $1mole$ of ammonia $(N{H_3})$ occupies $22.4litres$ and it has Avogadro’s molecules ($6.023 \times {10^{23}}$ molecules).
Therefore, $1litre$ contain $\dfrac{1}{{22.4}}$ moles and $10litres$ contain $\dfrac{{10}}{{22.4}}$ moles of ammonia.
Now, the number of molecules in $\dfrac{{10}}{{22.4}}$ moles of ammonia is $\dfrac{{10}}{{22.4}} \times 6.022 \times {10^{23}}$ molecules, but our aim to calculate the number of atom. In ammonia molecules there are four atoms. So multiply the number of molecules by four.
Number of atom is $10litres$ of ammonia $ = 4 \times \dfrac{{10}}{{22.4}} \times 6.022 \times {10^{23}}atoms$
On solving the above equation, we get
$ = 1.786 \times 6.022 \times {10^{23}}atoms$
Hence $10litres$ of ammonia contain $1.786 \times 6.022 \times {10^{23}}atoms$ . so the correct option is \[D.\]
Additional information:Avogadro’s number: Avogadro’s number of atom or molecules in mole of substance is equal to $6.023 \times {10^{23}}$. we can calculate the Avogadro’s number by dividing the charge of one mole of electron by the charge on a single electron we get a value of Avogadro’s number of $6.023 \times {10^{23}}$ particles per mole. The unit of avogadro’s number is $mo{l^{ - 1}}$ .
Ammonia: Ammonia is a colourless gas with a distinct odour composed of nitrogen and oxygen. The chemical formula of ammonia is $N{H_3}$ . Ammonia is an important component of the metabolic process.
Note: It is to be noted that one mole of any gas contains $22.4L$ volume at standard temperature and pressure of gas. We can calculate the volume at $STP$ by using the ideal gas equation. The ideal gas reaction is $PV = nRT$ .
Complete answer:
We know any type of gas occupies approximately $22.4litres$ of space at standard temperature pressure, we also know $1mole$ of contain Avogadro’s number of molecules.
So, we can say $1mole$ of ammonia $(N{H_3})$ occupies $22.4litres$ and it has Avogadro’s molecules ($6.023 \times {10^{23}}$ molecules).
Therefore, $1litre$ contain $\dfrac{1}{{22.4}}$ moles and $10litres$ contain $\dfrac{{10}}{{22.4}}$ moles of ammonia.
Now, the number of molecules in $\dfrac{{10}}{{22.4}}$ moles of ammonia is $\dfrac{{10}}{{22.4}} \times 6.022 \times {10^{23}}$ molecules, but our aim to calculate the number of atom. In ammonia molecules there are four atoms. So multiply the number of molecules by four.
Number of atom is $10litres$ of ammonia $ = 4 \times \dfrac{{10}}{{22.4}} \times 6.022 \times {10^{23}}atoms$
On solving the above equation, we get
$ = 1.786 \times 6.022 \times {10^{23}}atoms$
Hence $10litres$ of ammonia contain $1.786 \times 6.022 \times {10^{23}}atoms$ . so the correct option is \[D.\]
Additional information:Avogadro’s number: Avogadro’s number of atom or molecules in mole of substance is equal to $6.023 \times {10^{23}}$. we can calculate the Avogadro’s number by dividing the charge of one mole of electron by the charge on a single electron we get a value of Avogadro’s number of $6.023 \times {10^{23}}$ particles per mole. The unit of avogadro’s number is $mo{l^{ - 1}}$ .
Ammonia: Ammonia is a colourless gas with a distinct odour composed of nitrogen and oxygen. The chemical formula of ammonia is $N{H_3}$ . Ammonia is an important component of the metabolic process.
Note: It is to be noted that one mole of any gas contains $22.4L$ volume at standard temperature and pressure of gas. We can calculate the volume at $STP$ by using the ideal gas equation. The ideal gas reaction is $PV = nRT$ .
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