Question

Under the identical conditions a gas cxhy diffuses two times slower than helium what is the value ofx+y

Answer 1

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Answer 2

Given: Under identical conditions a gas $C_xH_y$ diffuses two times slower than helium.

To find: We have to find the value of x+y.

Solution:

According to the law of diffusion rate of diffusion is inversely proportional to the mass of the species.

Let the rate of diffusion of $C_xH_y$ be x and He is y.

The molecular weight of He is 4.

The molecular weight of $C_xH_y$ is 12x+y.

So, we can write-

$\frac{x}{y} = \sqrt{ \frac{4}{12x + y} }$

Given that the unknown gas diffuses two times slower than He.

So, x/y=1/2

$\frac{1}{2} = \sqrt{ \frac{4}{12x + y} } \\ \frac{1}{4} = \frac{4}{12x + y} \\ 12x + y = 16$

Thus 12x+y=16

Now methane can have a molecular weight of 16.

So, the value of x and y is 1 and 4.

Thus x+y=1+4

x+y=5

The value of x+y is 5.