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Graham's Law of Effusion

Two gases NH ...

Question

Two gases NH₃ and HBr are introduced simultaneously into opposite ends of an open tube 1m long. The distance at which a ring of NH₄Br is formed from the NH₃ end is

73.33 cm

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60.6 cm

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54.75 cm

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68.55 cm

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Solution

The correct option is D 68.55 cm
Let the distance at which a ring of N $H_{4}$ Br is formed from N $H_{3}$ end = 'x' cm.
Thus from HBr end it is = (100 - x) cm
According to Grahm's law of diffusion of gases,
Rate of diffusion, r $\propto \frac{1}{\sqrt{M}}$
$\Rightarrow \frac{r_{1}}{r_{2}} = \frac{x}{(100 - x)} = \sqrt{\frac{M_{H B r}}{M_{N H_{3}}}}$
$\Rightarrow \frac{x}{(100 - x)} = \sqrt{\frac{81}{17}} = 2.18$
$\Rightarrow$ x = 218 - 2.18x
x = 68.55 cm.

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Two gases NH3 and HBr are introduced simultaneously into opposite ends of an open tube 1m long. The distance at which a ring of NH4Br is formed from the NH3 end is

Two gases NH₃ and HBr are introduced simultaneously into opposite ends of an open tube 1m long. The distance at which a ring of NH₄Br is formed from the NH₃ end is