Question

A gas jar contains $7.2 \times {10^{20}}$ molecules of ammonia gas. Find the number of moles, weight in grams, volume in $c{m^3}$ of ammonia gas at S.T.P. [N=14, H=1].

Answer 1

The formula to find the number of moles is
${\text{Number of moles = }}\dfrac{{{\text{Weight}}}}{{{\text{Molecular weight}}}}$

Complete Step-by-Step Solution:
We will find the given quantities one by one.
- The atomic weights of nitrogen and hydrogen atoms are given in the molecule.
We can write that
Molecular weight of $N{H_3}$ = Atomic weight of N + 3(Atomic weight of H) = 14 + 3(1) = 17$gmmo{l^{ - 1}}$.
- Now, we know that one mole of any substance contains $6.022 \times {10^{23}}$ number of molecules or atoms.
Here, the gas contains 7.2$\times {10^{20}}$ molecules.
So, we can write that the moles of the gas = $\dfrac{{{\text{Number of molecules}}}}{{{\text{Avogadro number}}}}$ =$\dfrac{{7.2 \times {{10}^{20}}}}{{6.022 \times {{10}^{23}}}}$ =$1.1956 \times {10^{ - 3}}$
Now, we need to find the weight of the given molecules of the gas in gms.
- We know that the number of moles of the gas and weight of the gas are related by the following formula.
${\text{Number of moles = }}\dfrac{{{\text{Weight}}}}{{{\text{Molecular weight}}}}$
Putting the available values in the given equation, we get
$1.1956 \times {10^{ - 3}} = \dfrac{{{\text{Weight}}}}{{17}}$
So, weight = $1.1956 \times {10^{ - 3}} \times 17$ = 0.03325 gm
Now, we will find the volume of the given gas.
- Note that any gas whose concentration is 1 mole at S.T.P. has a volume of 22.4 L.
But we need to give the volume in $c{m^3}$ units.
We know that $1{\text{ L = 1000c}}{{\text{m}}^3}$
So, 22.4 L = $22.4 \times 1000 = 22400c{m^3}$
Now, we can write that
At STP, 1 mole of any gas has a volume of 22400$c{m^3}$,
So, $1.1956 \times {10^{ - 3}}$ moles of gas will have volume = $1.1956 \times {10^{ - 3}} \times 22400 = 26.7814c{m^3}$

Thus, we can conclude that the number of moles of the gas is $1.1956 \times {10^{ - 3}}$, Weight of the gas = 0.03325gm and Volume of the gas = 26.7814$c{m^3}$

Note: Always remember the relation between the different units of volume. Some of them are given as below.
$1{m^3} = 1000L$
$1000c{m^3} = 1L$
$1c{m^3} = 1mL$