Question

Volume of $1$ equivalent of ${{O}_{2}}$ at STP is
(A) $5.6L$
(B) $11.2L$
(C) $22.4L$
(D) $20L$

Answer 1

We know that using the Avogadro’s law which states $1$ mol of gas at NTP occupies 22.4 L of the gas, we can easily find the volume occupied by the oxygen atom at NTP but we have to find out the moles of oxygen from the moles of ${{O}_{2}}$ and moles of ${{O}_{2}}$ .

Complete answer:
This is based on Avogadro's law. So, let’s discuss Avogadro's law. This law states that any gas molecule at the standard conditions of temperature(i.e. ${}^\circ C$ ) and pressure(i.e. $1$ atm) contains Avogadro number of particles and $1$ mol of the gas at STP or NTP occupies the volume of litres. Avogadro’s law is applicable only at the conditions of standard temperature and pressure and if the pressure and temperature are constant, then on increasing the amount of the gas, its volume also increases and vice-versa
It is already known that $1$ mole of the gas( or $32g$ of ${{O}_{2}}$ ) is equivalent to $22.4$ Litres of the oxygen gas. For finding the gram equivalent volume of oxygen( O2), we equate the following: $32g$ of ${{O}_{2}}\to 22.4L$ of ${{O}_{2}}$ . So, $8g$ is equivalent to $=\left( \dfrac{22.4}{32} \right)\times 8=5.6\text{ }L$ of the gas. Thus, the gram equivalent volume of oxygen is $5.6L.$
Therefore, the correct answer is option A.

Note:
Remember that STP is mostly used to calculate the gas density. The term which is similar to STP is NTP. The abbreviation of NTP is normal temperature and pressure. It defines temperature at and pressure at one atmosphere. NTP is set but mostly is considered as temperature. STP is mainly used to express the fluid flow.