Question

The average molecular mass of a mixture of gas containing nitrogen and carbon dioxide is $36$. The mixture contains $280 g$ of nitrogen. Therefore, the amount of $C{{O}_ {2}} $ present in the mixture is:
A) 440 g
B) 44 g
C) 4.4 g
D) 880 g

Answer 1

Hint: The amount of substance in a given sample of matter is defined as the number of discrete atomic-scale particles in it divided by the Avogadro constant NA. In a truly atomistic view, the amount of substance is simply the number of particles that constitute the substance.

Complete step by step answer:
We have been provided with the average molecular mass of a mixture of gas containing nitrogen and carbon dioxide: 36.
The number average molecular mass is the ordinary arithmetic mean or average of the molecular masses of the individual macromolecules. It is determined by measuring the molecular mass of n polymer molecules, adding the masses, and dividing by n.
So, let us take the moles of carbon dioxide gas be x,
Now, the moles of ${{N}_ {2}} $ gas: $\dfrac {280}{28} =10$, (the molar mass of ${{N}_ {2}} $ is 28),
Also, the molecular mass of carbon dioxide gas:44,
So, the average molecular mass of mixture: $\left (10\times 28 \right) +x+\dfrac {44}{10} +x=36$,
Solving the above equation:
\[280+\dfrac{44x}{10} +x=36\],
 Simplifying it further:
     \[280+44x=360-36x\]
Noe, separating the variable and constant terms:
     \[44x-36x=360-280\],
So, from this the value of x: $8x=80$,
So, the average molecular mass comes out to be: $x=10$ moles.
So, the amount of carbon dioxide gas: $10\times 44=440$ g.
Therefore, we can say that option (A) is correct.

Note: The mole was defined in such a way that the molar mass of a compound, in g/mol, is numerically equal (for all practical purposes) to the average mass of one molecule, in Daltons. Thus, for example, the average mass of a molecule of water is about 18.0153 Daltons, and the molar mass of water is about 18.0153 g/mol.