Question: What is the volume occupied by 4 grams of ethylene at STP( in litres)? A.22.4 B.11.2 C.5.6 D...

Question

What is the volume occupied by 4 grams of ethylene at STP( in litres)?
A.22.4
B.11.2
C.5.6
D.3.2

Answer 1

Solution

Try to find out the moles of ethylene when 7 grams of it is given in STP condition. Now, use the condition of STP and with the help of a number of moles, find out the volume occupied by 7 grams of it.

Complete answer:
In order to answer our question, we need to know what STP, NTP is and how molar mass plays a role in it. Now, the full form of STP is standard temperature and pressure and NTP stands for normal temperature and pressure. These are certain conditions that are set by IUPAC for ease. In case of STP, the temperature is taken to be ${{0}^{o}}C$ and the pressure is taken to be 100kPa, whereas in the case of NTP, the temperature is taken to be ${{20}^{0}}C$ and the pressure is 1 bar.
Generally, when a gas is present at STP, then it means that if the gas occupies one mole, then in the given conditions, it will also occupy a volume of 22.4 litres. For example, oxygen has a molar mass of 32, as it is a diatomic molecule. So, when one mole of oxygen molecule, i.e 32 grams of oxygen is present in the conditions of STP, then the volume occupied by it is exactly 22.4 litres. Now, with the help of this, we will try to solve our question.
Now, we have been given ethylene. Ethylene has the molecular formula ${{C}_{2}}{{H}_{2}}$ and has a molar mass of $28g\,mo{{l}^{-1}}$ . So, when 28 grams of ethylene is present in STP, then a volume of 22.4 litres is occupied by it. However, we have been given that only 4 grams of ethylene is present. So, we have:
$\begin{aligned} & 28grams\,{{C}_{2}}{{H}_{2}}\,gives\,22.4\,litres \\\ & 7\,grams{{C}_{2}}{{H}_{2}}\,gives\,(\dfrac{7}{28})\times 22.4\,litres=3.2\,litres \\\ \end{aligned}$
So, 7 grams of ethylene occupies a volume of 3.2 litres in standard temperature and pressure conditions.

So, we get option D as our correct answer.

Note:
It is to be noted that the STP and NTP conditions are a sole result and are obtained from the ideal gas equation, that is $PV=nRT$ . It’s only that STP and NTP fixes the values of pressure and temperature. As the number of moles and R is always constant, the volume can be calculated.