Question

If a gas effuses one third as fast a methane ,what is the molecular weight of A?



Plz help me ......​

Answer 1

Answer:Explanation:

rate of effusion is inversely related to root of molar mass

therefore 1/3=(m/16)^1/2

therefore molecular weight is 16/9

Answer 2

Concept:

The amount of molecules that diffuse through the hole in a given amount of time is known as the rate of effusion. The effusion rate is inversely proportional to the square root of the molecular weight under constant pressure and temperature.

Given:

The unknown gas effuses one-third as fast as methane = $\frac{1}{3}$

Find: If a gas effuses one-third as fast as methane, what is the molecular weight of A?

Solution:

According to Graham's law, the rate of diffusion is inversely proportional to the square roots of its molecule weight.

Let m be the mass of the unknown gas A.

m₂ is the mass of methane gas.

The molecular mass of methane gas = 16 g/mol

r₁  = rate of diffusion of the unknown gas A.

r₂ = rate of diffusion of methane gas.

$\frac{r_{1} }{r_{2} }$ = $\sqrt{\frac{m_{2} }{m_{1} } }$

It is given that $\frac{r_{1} }{r_{2} }$ = $\frac{1}{3}$

Thus, $\frac{1}{3}$ = $\sqrt{\frac{m_{2} }{m_{1} } }$

$\frac{1}{3}$ = $\sqrt{\frac{16}{m_{1} } }$

Squaring both sides and we get,

$\frac{1}{9}$ = $\frac{16 }{m_{1} }$

m₁ = 16 * 9

m₁ = 144 g/mol

Therefore, the molecular mass of unknown gas A is 144 g/mol.

Hence, if a gas effuses one-third as fast as methane, the molecular weight of A is 144 g/mol.

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