Question

What is the molecular weight of a gas that diffuses 25 times as fast as helium?

Answer 1

Answer : -

64.0 g/mol

Explanation:

64.0 g/mol is the molecular weight of a gas that diffuses 0.25 times as fast as helium.

Answer 2

Given:  A gas X which diffuses 25 times as fast as Helium

To Find: Molecular Weight of Gas X

Solution:

• According to Graham's Law of Diffusion,

                rate of diffusion of gas ∝ $\frac{1}{\sqrt{M} }$            (1)

where M is molecular weight of gas.

• Now as per the question, let rate of diffusion of Helium be $r_{1}$ and let rate of diffusion of gas X of molecular mass X be $r_{2}$.
• ATQ $r_{2}$ = 25 $r_{1}$
• Substituting these values in (1) we get:

                     $\frac{25 r_{1} }{r_{1} }$ = $\frac{\sqrt{4} }{\sqrt{M} }$

               ⇒ 25 = $\frac{2}{\sqrt{M} }$

              ⇒ $\sqrt{M}$ = $\frac{2}{25}$

             ⇒ M = $\frac{4}{625}$ = 0.0064 g                 (squaring both sides)

Hence molecular weight of gas is 0.064 g.