Question

The total weight of a gas mixture which contains 11.2 litres of \[C{O_2}\] , 5.6 litre of \[{O_2}\] , 22.4 litres of \[{N_2}\] , 44.8 litres of Heat \[0^\circ C\] and 760 mm of Hg pressure is:
A.44 g
B.22g
C.33 g
D.66 g

Answer 1

Hint: To solve this question, we must first calculate the number of moles of each gas present using the volume of the gas. The, we must find the weight of individual gases in the given mixture. We must add these values to get the final weight of the mixture.

Complete Step-by-Step Answer:
Before we move forward with the solution of the given question, let us first discuss some important basic concepts.
Mole concept: Moles are used to measure the amount of matter present in a given sample of a substance. To be more specific, it helps in measuring the number of atoms or molecules present in the given sample of the substance. One mole of any substance contains \[6.022 \times {10^{23}}\] atoms or molecules.
This means that 1 mole of any gas is 22.4 litre when the external physical conditions are temperature = 273.15 K and pressure = 1 atm.
The data that has been given to us is:
Temperature = \[0^\circ C\] = 273.15 K,
Pressure = 760 mm of Hg = 1 atm
Volume of \[(C{O_2})\] = 11.2 litres
Volume of \[({O_2})\] = 5.6 litres
Volume of \[({N_2})\] = 22.4 litres
Volume of (He) = 44.8 litres
Hence, the physical conditions are STP in the given question. Hence the volume of any gas at these conditions is 22.4 L. Hence the number of moles of the given gases can be calculate as:
Number of moles of \[(C{O_2})\] = 11.2 litres/22.4 L = 0.5 moles
Number of moles of \[({O_2})\] = 5.6 litres/22.4 L = 0.25 moles
Number of moles of \[({N_2})\] = 22.4 litres/22.4 L = 1 mole
Number of moles of (He) = 44.8 litres/22.4 L = 2 moles
The molar masses of these gases are \[C{O_2}\] = 44 g/mol, \[{O_2}\] = 32 g/mol, \[{N_2}\] = 28 g/mol, He = 2 g/mol
The weight of these gases can be calculated using the formula:
Weight of the gas = (no. of moles) (molar mass)
Hence the total volume of the system can be calculated as:
 \[W = W(C{O_2}) + W({O_2}) + W({N_2}) + W\left( {He} \right)\]
W = 22 + 8 +28 + 8
W = 66 g
Hence, Option D is the correct option

Note: Avogadro’s Hypothesis states that equivalent volumes of all gases contain equal numbers of particles at constant values of temperature and pressure. Stand Temperature Pressure or STP can be defined as a physical condition where the temperature of the system is 273.15 K and the surrounding pressure is 1 atm. Molar volume of a given gas can be defined as the volume of a given gas at STP. The volume of any gas at STP, containing one mole of the gas is 22.4 litres.