Question

56. If the molecules of so2 effuses with a speed of 150cm in certain period of time, in
travelled by the molecules of CH4, effusing in the same time is
1) 300 cm
2 600 cm
3) 37.5 cm
4 75 cm​

Answer 1

Answer:

1) 300 cm

Explanation:

According to Graham’s law of effusion the rate of effusion of gasses is inversely proportional to the square root ratios of their molar masses.

formula

for 1 gas

     Rate of effusion ∝ 1 / √M

For 2 gasses A and B

     (Rate of effusion of A / Rate of effusion of B) =   (√Mᵇ / √Mᵃ)

given data

     Rate of effusion of SO₂ (B) = 150 cm

     Rate of effusion of CH₄ (A) = ?

As we know that

   molar mass of CH₄ (A) = 16.04 g/mol

   molar mass of SO₂ (B) = 64.006 g/mol

Put the values

   (Rate of effusion of CH₄ / 150 cm) =   (√64.006 / √16.04)

   (Rate of effusion of CH₄ / 150 cm) =   (8.004 / 4.004)

   (Rate of effusion of CH₄ / 150 cm) =   2

   Rate of effusion of CH₄  =   2 × 150 cm

   Rate of effusion of CH₄  =   300 cm

Answer 2

Given:

Rate of effusion of $SO_2$ = 150 cm

To find:

Rate of effusion of $CH_4$ = ?

Formula to be used:

(Rate of effusion of A / Rate of effusion of B) = ($\sqrt{M_B} /\sqrt{M_A}$

where 'M' is molar mass

Molar mass of $CH_4 and SO_2$ are 16.04 g/mol and 64.006 g/mol respectively.

Calculation:

Let gas A be $CH_4$ and gas B be $SO_2$

$(Rate of effusion of A / Rate of effusion of B) = ([tex]\sqrt{M_B} /\sqrt{M_A}$

$(Rate of effusion of CH_4 / 150 cm) = (\sqrt{64.006} /\sqrt{16.04} )$

$(Rate of effusion of CH_4 / 150 cm) = (8.004/4.004)$

$(Rate of effusion of CH_4 / 150 cm) = 2$

$Rate of effusion of CH_4  = 2\times 150$

$Rate of effusion of CH_4 = 300cm$

Conclusion:

The rate of effusion of $CH_4$ is 300cm

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