
For 10 minute each \[{0^o}C\], at, from two identical holes nitrogen and an unknown gas are leaked into a common vessel of 4 litre capacity. The resulting pressure is 2.8 atm and the mixture contains a 0.4 mole of nitrogen. What is the molar mass of unknown gas?
Answer
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Hint: The general gas equation, also known as the ideal gas law, is the state equation of a potentially ideal gas. While it has many drawbacks, it is a reasonable approximation of the action of certain gases under several conditions.
Complete answer:
The molar mass of a chemical compound is known as the mass of a sample divided by the amount of material in the sample, measured in moles, in chemistry. The molar mass of a material is a bulk property, not a molecular property.
Let's call the unknown gas X.
Let \[{M_N}\] and \[{M_X}\] represent the nitrogen and other gas molar masses, respectively.
Using the ideal gas equation-
PV = nRT
Given data,
P = 2.8 bar
V = 4 L
T = 273 K
R = \[{\mathbf{0}}.{\mathbf{0821LatmJ}}{{\mathbf{K}}^{ - 1}}{\mathbf{mo}}{{\mathbf{l}}^{ - 1}}\]
The total number of moles in the solution (n) = \[ = \dfrac{{2.8 \times 4}}{{0.0821 \times 273}} = 0.5{\text{ moles }}\]
The number of moles of nitrogen gas is given as 0.4 moles
The number of moles of gas X = 0.5 – 0.4 = 0.1
Now, according to Graham's diffusion rule,
\[{\text{ rate of effusion }} = \dfrac{{{\text{ no}}{\text{. of moles }}}}{{{\text{ time taken }}}}\]
Given that t=10 minutes
Rate of effusion of nitrogen \[\left( {{{\mathbf{r}}_{\text{N}}}} \right) = \dfrac{{0.4}}{{10}} = 0.04\]
Rate of effusion of gas X \[\left( {{{\text{r}}_{\text{X}}}} \right) = \dfrac{{0.1}}{{10}} = 0.01\]
Furthermore, we are aware that
\[\dfrac{{{{\text{r}}_{\text{N}}}}}{{{{\text{r}}_{\text{X}}}}} = \sqrt {\dfrac{{{{\text{M}}_{\text{X}}}}}{{{{\text{M}}_{\text{N}}}}}} \]
We know that the molar mass of nitrogen gas is 28g.
\[\therefore \dfrac{{{\mathbf{0}}.{\mathbf{04}}}}{{{\mathbf{0}}.{\mathbf{01}}}} = \sqrt {\dfrac{{{{\mathbf{M}}_{\mathbf{X}}}}}{{{\mathbf{28}}}}} \]
\[ \Rightarrow \sqrt {{{\text{M}}_{\text{X}}}} = 4 \times \sqrt {28} \]
Squaring on both sides, we have
\[{{\mathbf{M}}_{\mathbf{X}}} = {\mathbf{16}} \times {\mathbf{28}} = {\mathbf{448gm}}/{\mathbf{mol}}\]
Therefore the molar mass of unknown gas is 448gm/mol.
Note:
Nitrogen is an inert gas, which means it doesn't react chemically with other gases and isn't harmful. However, breathing pure nitrogen is lethal. This is due to the gas's ability to displace oxygen in the lungs. According to the US Chemical Safety and Hazard Investigation Board, unconsciousness can occur in as few as one or two breaths.
Complete answer:
The molar mass of a chemical compound is known as the mass of a sample divided by the amount of material in the sample, measured in moles, in chemistry. The molar mass of a material is a bulk property, not a molecular property.
Let's call the unknown gas X.
Let \[{M_N}\] and \[{M_X}\] represent the nitrogen and other gas molar masses, respectively.
Using the ideal gas equation-
PV = nRT
Given data,
P = 2.8 bar
V = 4 L
T = 273 K
R = \[{\mathbf{0}}.{\mathbf{0821LatmJ}}{{\mathbf{K}}^{ - 1}}{\mathbf{mo}}{{\mathbf{l}}^{ - 1}}\]
The total number of moles in the solution (n) = \[ = \dfrac{{2.8 \times 4}}{{0.0821 \times 273}} = 0.5{\text{ moles }}\]
The number of moles of nitrogen gas is given as 0.4 moles
The number of moles of gas X = 0.5 – 0.4 = 0.1
Now, according to Graham's diffusion rule,
\[{\text{ rate of effusion }} = \dfrac{{{\text{ no}}{\text{. of moles }}}}{{{\text{ time taken }}}}\]
Given that t=10 minutes
Rate of effusion of nitrogen \[\left( {{{\mathbf{r}}_{\text{N}}}} \right) = \dfrac{{0.4}}{{10}} = 0.04\]
Rate of effusion of gas X \[\left( {{{\text{r}}_{\text{X}}}} \right) = \dfrac{{0.1}}{{10}} = 0.01\]
Furthermore, we are aware that
\[\dfrac{{{{\text{r}}_{\text{N}}}}}{{{{\text{r}}_{\text{X}}}}} = \sqrt {\dfrac{{{{\text{M}}_{\text{X}}}}}{{{{\text{M}}_{\text{N}}}}}} \]
We know that the molar mass of nitrogen gas is 28g.
\[\therefore \dfrac{{{\mathbf{0}}.{\mathbf{04}}}}{{{\mathbf{0}}.{\mathbf{01}}}} = \sqrt {\dfrac{{{{\mathbf{M}}_{\mathbf{X}}}}}{{{\mathbf{28}}}}} \]
\[ \Rightarrow \sqrt {{{\text{M}}_{\text{X}}}} = 4 \times \sqrt {28} \]
Squaring on both sides, we have
\[{{\mathbf{M}}_{\mathbf{X}}} = {\mathbf{16}} \times {\mathbf{28}} = {\mathbf{448gm}}/{\mathbf{mol}}\]
Therefore the molar mass of unknown gas is 448gm/mol.
Note:
Nitrogen is an inert gas, which means it doesn't react chemically with other gases and isn't harmful. However, breathing pure nitrogen is lethal. This is due to the gas's ability to displace oxygen in the lungs. According to the US Chemical Safety and Hazard Investigation Board, unconsciousness can occur in as few as one or two breaths.
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