Question

Ratio of active masses of ${\text{22 g}}$ of ${\text{C}}{{\text{O}}_{\text{2}}}$, ${\text{3}}\,{\text{g}}$ of ${{\text{H}}_{\text{2}}}$and,${\text{7}}\,{\text{g}}$ of ${{\text{N}}_{\text{2}}}$ in a gaseous mixture:

A.$22:3:7$
B.$0.5:3:7$
C.$1:3:1$
D. $1:3:0.5$

Answer 1

Hint:: We should know about active mass to determine the ratio of active masses. We should also know to calculate the molar mass and mole. We can determine the mole of a compound by using the mole formula. The ratio gives the comparative amount of one concerning another.

Formula used: ${\text{Mole}}\,{\text{ = }}\,\dfrac{{{\text{Mass}}}}{{{\text{Molar}}\,{\text{mass}}}}$


Complete step by step solution:The molar concentration is known as active mass. Molar concentration means gram-mole amount present in per liter volume. Here, the volume is not given so, we assume the volume of the mixture is V.

To take the ratio of molar concentration we will first determine molar concentration by dividing the mole of each with volume. Since volume is the same for all species, we can take the ratio of the number of moles of each species.

The mole is determined by dividing the mass of a compound by its molar mass.

${\text{Mole}}\,{\text{ = }}\,\dfrac{{{\text{Mass}}}}{{{\text{Molar}}\,{\text{mass}}}}$

The molar mass of carbon dioxide is $44\,{\text{g/mol}}$..

Substitute $44\,{\text{g/mol}}$ for molar mass and ${\text{22g}}$ for mass of carbon dioxide.

$\,{\text{mol}}\,\,{\text{ = }}\,\dfrac{{{\text{22}}\,{\text{g}}}}{{44\,{\text{g/mol}}}}$

${\text{Moles}}\,{\text{of C}}{{\text{O}}_2}\,{\text{ = }}\,0.5$

So, ${\text{22 g}}$ carbon dioxide is equal to $0.5$ mole of carbon dioxide.
Molar mass of the hydrogen is $2.016\,{\text{g/mol}}$.

Substitute $2.016\,{\text{g/mol}}$ for molar mass and $3$ gram for mass of hydrogen.

${\text{mole}}\,{\text{ = }}\,\dfrac{{{\text{3}}\,{\text{g}}}}{{2.016\,{\text{g/mol}}}}$

${\text{mole}}\,{\text{of }}{{\text{H}}_2}\,{\text{ = }}\,1.5\,$

So, $3$ gram hydrogen is equal to $1.5$mole of hydrogen.

Molar mass of the nitrogen is $28\,{\text{g/mol}}$.

Substitute $28\,{\text{g/mol}}$ for molar mass and $7$ gram for mass of nitrogen.

${\text{mole}}\,{\text{ = }}\,\dfrac{{7\,{\text{g}}}}{{28\,{\text{g/mol}}}}$

${\text{mole}}\,{\text{of }}{{\text{N}}_2}\,{\text{ = }}\,0.25$

So, $7$ gram nitrogen is equal to $0.25$mole of nitrogen.

Ratio of a compound is determined by dividing the mole of that compound by the total mole.

The total moles is,

$ = 0.5\, + \,1.5 + 0.25$

$ = 2.25$

The mole ratio of carbon dioxide is,

$ = \dfrac{{0.5}}{{2.25}}$

$ = 0.2$

The mole ratio of hydrogen is,

$ = \dfrac{{1.5}}{{2.25}}$

$ = 0.6$

The mole ratio of nitrogen is,

$ = \dfrac{{0.25}}{{2.25}}$
$ = 0.1$

We will multiply the ratio with $5$ to convert into the whole number and take their ratio.

\[ = 0.2 \times 5:0.6 \times 5:0.1 \times 5\]

\[ = 1:3:0.5\]

So, the ratio of active masses of ${\text{22 g}}$ of ${\text{C}} {{\text{O}}_{\text{2}}}$,${\text{3}}\ {\text{g}}$ of ${{\text{H}}_{\text{2}}}$ and ,${\text{7}}\,{\text{g}}$ of ${{\text{N}}_{\text{2}}}$ in a gaseous mixture is $1:3:0.5$.


Therefore, option (D) $1:3:0.5$ is correct.


Note: Molar concentration is known as molarity. Molarity is defined as the mole of solute present in per litter of the solution. Molarity is directly proportional to density. In the case of solid and liquid, the value of active mass is taken as one. Solid and liquid do not dissociate in solution, so they are considered as one unit. The compound which dissociates produces ions, so their concentration varies in solid and solution. The active mass is the amount that reacts in a reaction.