Question

5.6litre of oxygen at STP is equivalent to:
A. \[1mole\]
B. \[\dfrac{1}{2}mole\]
C. \[\dfrac{1}{4}mole\]
D. \[\dfrac{1}{8}mole\]

Answer 1

Hint: The abbreviated form of STP is Standard Temperature and Pressure. It defines the temperature at zero degrees Celsius and pressure at one atmosphere. Zero degree Celsius can also be expressed in kelvin as \[273K\].

Complete step by step answer:
Generally, it is known that at STP, the gases have a volume of \[22.4L\] per mole. Thus, it is clear that one mole of gas contains \[22.4L\]
If the gas contains one mole, then at STP, its volume is \[22.4L\].
Thus, \[1mole\] of gas has \[22.4L\] volume present in it.
The litre of oxygen present at STP on \[\dfrac{1}{2}mole\] can be calculated as,
\[ \Rightarrow \dfrac{1}{2} \times 22.4L\]
\[ \Rightarrow 11.2L\]
Thus, \[11.2L\] of oxygen at STP is equivalent to \[\dfrac{1}{2}mole\]
 The litre of oxygen present at STP on \[\dfrac{1}{4}mole\] can be calculated as,
\[ \Rightarrow \dfrac{1}{4} \times 22.4L\]
\[ \Rightarrow 5.6L\]
Thus, \[5.6L\] of oxygen at STP is equivalent to \[\dfrac{1}{4}mole\]
The litre of oxygen present at STP on \[\dfrac{1}{8}mole\] can be calculated as,
\[ \Rightarrow \dfrac{1}{8} \times 22.4L\]
\[ \Rightarrow 2.8L\]
Thus, \[2.8L\] of oxygen at STP is equivalent to \[\dfrac{1}{8}mole\]
Thus, \[5.6litre\] of oxygen at STP is equivalent to \[\dfrac{1}{4}mole\]

So, the correct answer is Option C.

Note: 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 the temperature at \[{25^\circ }C\] and pressure at one atmosphere. NTP is set as \[101.325kPa\] but mostly \[{25^\circ }C\] is considered as temperature. STP is mainly used to express the fluid flow rate and the volume of gases which are dependent on temperature and pressure. NTP is mainly used as the standard reference for testing and documentation of fan capacities. One mole of a gas at NTP occupies \[22.4L\]. For oxygen atoms, one mole at NTP is equal to \[11.2L\].