what is the molecular weight of an unknown gas if 200ml of this gas diffuses through an apparatus in 180 secs while 250mL of NO2 under the same conditions diffuses through the same apparatus in 170 secs
a. 8.97 g/mol
b. 60.88 g/mol
c. 80.52 g/mol
805.20 g/mol
The Correct Answer and Explanation is :
To find the molecular weight of the unknown gas, we can use Graham’s law of diffusion, which relates the rates of diffusion of two gases to their molar masses. According to Graham’s law:
[
\frac{\text{Rate of diffusion of Gas 1}}{\text{Rate of diffusion of Gas 2}} = \sqrt{\frac{M_2}{M_1}}
]
Where:
- ( M_1 ) and ( M_2 ) are the molar masses of the two gases.
- Rate of diffusion is inversely proportional to the time taken for the gas to diffuse through the apparatus.
Let’s apply this to the problem:
- Gas 1 is the unknown gas, and Gas 2 is ( \text{NO}_2 ).
- The volume of the unknown gas is 200 mL, and it takes 180 seconds to diffuse.
- The volume of ( \text{NO}_2 ) is 250 mL, and it takes 170 seconds to diffuse.
First, calculate the rate of diffusion for both gases:
[
\text{Rate of diffusion} = \frac{\text{Volume}}{\text{Time}}
]
For the unknown gas:
[
\text{Rate of diffusion of unknown} = \frac{200}{180} = 1.11 \, \text{mL/s}
]
For ( \text{NO}_2 ):
[
\text{Rate of diffusion of NO}_2 = \frac{250}{170} = 1.47 \, \text{mL/s}
]
Now, applying Graham’s law:
[
\frac{1.11}{1.47} = \sqrt{\frac{M_{\text{NO}2}}{M{\text{unknown}}}}
]
Since the molar mass of ( \text{NO}2 ) is 46.0055 g/mol, we substitute into the equation: [ \frac{1.11}{1.47} = \sqrt{\frac{46.0055}{M{\text{unknown}}}}
]
Squaring both sides:
[
\left( \frac{1.11}{1.47} \right)^2 = \frac{46.0055}{M_{\text{unknown}}}
]
Solving for ( M_{\text{unknown}} ):
[
M_{\text{unknown}} = \frac{46.0055}{\left( \frac{1.11}{1.47} \right)^2}
]
[
M_{\text{unknown}} = \frac{46.0055}{0.514} \approx 89.5 \, \text{g/mol}
]
Since 80.52 g/mol is the closest option, the molecular weight of the unknown gas is 80.52 g/mol.
Thus, the correct answer is (c) 80.52 g/mol.