If a sample of O₂ gas (2.0 mmol) effused through a pinhole in 5.0 s, it will take 5.9 s for the same amount of CO₂ to effuse under the same conditions.
A sample of O₂ gas (2.0 mmol) effused through a pinhole in 5.0 s. The rate of effusion of O₂ is:
[tex]rO_2 = \frac{2.0mmol}{5.0s} = 0.40mmol/s[/tex]
Knowing their molar masses (M), we can calculate the rate of effusion of CO₂ using Graham's law.
[tex]\frac{rCO_2}{rO_2} = \sqrt{\frac{M(O_2)}{m(CO_2)} } \\\\\frac{rCO_2}{(0.40mmol/s)} = \sqrt{\frac{(32.00g/mol)}{(44.01g/mol) }} \\\\rCO_2 = 0.34 mmol/s[/tex]
The rate of effusion of CO₂ is 0.34 mmol/s. The time for 2.0 mmol to effuse is:
[tex]2.0 mmol \times \frac{1s}{0.34mmol} = 5.9 s[/tex]
If a sample of O₂ gas (2.0 mmol) effused through a pinhole in 5.0 s, it will take 5.9 s for the same amount of CO₂ to effuse under the same conditions.
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The time taken for the same amount of CO₂ to effuse under the same condition is 5.9 s
Let t₁ be the time taken for O₂ to effuse
Let M₁ be thr molar mass of O₂
Let t₂ be the time taken for CO₂ to effuse
Let M₂ be the molar mass of CO₂
Time taken for O₂ to effuse (t₁) = 5 s
Molar mass of O₂ (M₁) = 2 × 16 = 32 g/mol
Molar mass of CO₂ (M₂) = 12 + (2×16) = 44 g/mol
Time taken for CO₂ to effuse (t₂) = ?
[tex]\frac{t_{2}}{t_{1}} = \sqrt{\frac{M_{2}}{M_{1}}} \\\\\frac{t_{2}}{5} = \sqrt{\frac{44}{32}[/tex]
Cross multiply
[tex]t_{2} = 5 \sqrt{\frac{44}{32}[/tex]
Therefore, it will take 5.9 s for the same amount of CO₂ to effuse under the same condition.
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