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Graham's Law of Effusion

A small quant...

Question

A small quantity of gaseous $N H_{3}$ and HBr are introduced simultaneously into the opposite ends of an open tube that is 1m long. Calculate the distance of the white solid $N H_{4} B r$ formed from the end that was used to introduce $N H_{3}$ .

A

63.46 cm

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B

68.55 cm

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C

74.21 cm

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D

76.09 cm

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Solution

The correct option is B
68.55 cm

By Graham's law of diffusion:

$\frac{r_{N H_{3}}}{r_{H B r}} = \sqrt{\frac{81}{17}} = 2.18$

Thus , $N H_{3}$ travels 2.18 times faster than HBr.In other words, $N H_{3}$ will travel 2.18 cm in the same time in which HBr travels 1cm.Given that the length of the tube=100cm.

Therefore, distance travelled in the tube by

$N H_{3} = \frac{2.18}{⟮ 2.18 + 1 ⟯} \times 100$ = 68.55 cm

Thus , $N H_{4} B r$ will first appear at a distance of 68.55 cm from $N H_{3}$ end.

Required distance = 68.55 cm from $N H_{3}$ end.

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