Question

The total volume of 2 gm of Helium and 7gm of nitrogen under STP conditions is:
A.22.4L
B.11.2L
C.16.8L
D.5.6L

Answer 1

Hint: STP stands for Standard Temperature and Pressure. The conditions of temperature and pressure are $0^\circ $Celsius and 1 atmospheric pressure. Such measures are used for the comparison basis and the data is used in that only.
The International Union of Pure and Applied Chemistry (IUPAC) has given a certain set of rules in order to determine the observations under the standard set of conditions.

Complete step by step answer:
2 gm Helium and 7gm Nitrogen are given, and under the STP the number of moles present in a gas or molecule is equal to its atomic mass. Thus the atomic mass of Helium is 4 g and it will have the number of molecules in it to 1 mole, similarly under the same set of conditions 28 g of Nitrogen will have 1 mole of particles.
For Helium 4 g have 1 mole of molecules, thus 2 g will have:
$= \dfrac{{2 \times 1}}{4}$
= 0.5moles
 Similarly 28 g of Nitrogen will have:
$ = \dfrac{{7 \times 1}}{{28}} $
= 0.25moles

Thus the total moles of Gases present in the reaction mixture at STP is:
= (0.5+0.25) moles
=0.75 moles

Therefore at the STP conditions the gas occupies 22.4 L of volume thus;
At STP these gases will occupy$ = 0.75 \times 22.4L = 16.8L$

So, the correct answer is Option C.

Note: Initially at the STP temperature was 273.15 K and an absolute pressure of exactly 1 atm (101.325 kPa). But after the amendments, a temperature of 273.15 K and an absolute pressure of exactly 105 Pa (100 kPa, 1 bar) was made standard till date.