Question

Agas X diffuses through a pin hole at a rate 1/4th of the rate of diffusion of hydrogen gas under similar conditions, the gas is
O O₂
N2
O SO2
O CH4
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Answer 1

Given:

Agas X diffuses through a pin hole at a rate 1/4th of the rate of diffusion of hydrogen gas under similar conditions.

To find:

The gas X is ?

Calculation:

Let molecular weight of gas X is m :

According to GRAHAM'S LAW OF DIFFUSION:

$\therefore \: r \propto \dfrac{1}{ \sqrt{molecular \: mass} }$

$\implies \: \dfrac{r_{X} }{ r_{H_{2}} } = \dfrac{ \sqrt{ m_{H_{2}} } }{ \sqrt{ m_{X} } }$

$\implies \: \dfrac{ \frac{1}{4} \times r_{H_{2}} }{ r_{H_{2}} } = \dfrac{ \sqrt{ m_{H_{2}} } }{ \sqrt{ m_{X} } }$

$\implies \: \dfrac{1}{4} = \dfrac{ \sqrt{ m_{H_{2}} } }{ \sqrt{ m_{X} } }$

$\implies \: \dfrac{1}{4} = \dfrac{ \sqrt{2} }{ \sqrt{ m_{X} } }$

$\implies \: \dfrac{1}{16} = \dfrac{2}{ m_{X} }$

$\implies \: m_{X} = 32 \: g$

Now, since gas X has molecular mass 32 grams, then it has be to oxygen.

• This is because Oxygen gas molecule has 2 atoms each weighing 16 g. So, molecular weight of oxygen is 32 g.

X is Oxygen gas .

Answer 2

Answer:A

Explanation: