Question

1 $c{{m}^{3}}$ of ${{N}_{2}}O$ at S.T.P contains:

Answer 1

Hint: S.T.P means the standard temperature and pressure, and 1 mole of any gas will have a volume equal to 22400 $c{{m}^{3}}$. The number of moles in 1 $c{{m}^{3}}$ can be calculated by dividing the 1 by the S.T.P volume and in the number of molecules multiplying this value with the Avogadro's number.


Complete step by step answer:
- S.T.P means the gas is at standard temperature and pressure conditions, and the temperature taken in S.T.P is ${{0}^{\circ }}C$ or 273.015 K and the pressure taken at S.T.P is 1 bar.
- 1 mole of any gas will have a volume equal to 22400 $c{{m}^{3}}$ at S.T.P conditions, so 1 mole of ${{N}_{2}}O$ will also have a volume of 22400 $c{{m}^{3}}$ at S.T.P.
The number of moles in 1 $c{{m}^{3}}$ can be calculated by dividing the 1 by the S.T.P volume. So,
$\text{Moles}\ \text{in 1c}{{\text{m}}^{\text{3}}}=\dfrac{1}{22400}$
- We know that one mole of any substance will have $6.022\text{ x 1}{{\text{0}}^{23}}$ atoms, and this number is known as Avogadro’s number.
- To find the number of molecules in $1c{{m}^{3}}$, we multiply the number of moles of the gas given to the Avogadro’s number. On multiplying the value of Avogadro’s number, we get:
$No.\text{ of molecules = }\dfrac{6.023\text{ x 1}{{\text{0}}^{23}}}{22400}$
$=\dfrac{6.023\text{ x 1}{{\text{0}}^{23}}}{22400}=2.653\text{ x 1}{{\text{0}}^{19}}$
So, there are $2.653\text{ x 1}{{\text{0}}^{19}}$ molecules of ${{N}_{2}}O$ in 1 $c{{m}^{3}}$ at S.T.P.
- We have used no. of molecules because ${{N}_{2}}O$ is molecules and Avogadro's number is the same for all elements, molecules, and compounds or we can say that 1 mole of the atom will have Avogadro's number of atoms and 1 mole of molecules will have Avogadro's number of molecules.

Note: There is another term N.T.P which is used to define the condition of the gas, and its states that the gas is at normal temperature and pressure. The temperature taken is ${{20}^{\circ }}C$ and the pressure is 101.325 kPa.