Question

4 g of sulphur dioxide gas diffuse from a container in 8 minutes. Mass of helium gas diffusing in the same time interval

Answer 1

Answer:

Formula is $\frac{r_{SO_{2} } }{r_{He} }$=$\frac{V_{SO_{2} } }{Vx_{He} }$

$\frac{n_{SO2} }{n_{He} } =\sqrt{\frac{M_{He} }{M_{SO2} } }$

$\frac{\frac{4}{64} }{w/4} =\sqrt{\frac{4}{64} }$

$\frac{4*4}{64*w} =\frac{1}{4}$

weight/mass of helium=$\frac{4*4*4}{64} =\frac{64}{64}=1g$

Answer 2

mass of helium gas is 1g .

using Graham's law of diffusion,

rate of diffusion of gas is inversely proportional to square root of its molecular mass.

i.e., r ∝ 1/√M

and rate of diffusion = volume of gas diffused/time taken

from Avogadro's law, at same temperature and pressure, no of molecules of gas remains constant in the given container.

i.e., V ∝ n

so, n ∝ 1/√M

and hence, n1/n2 = √{M2/M1}

here, 4 g of sulphur dioxide gas diffuse from a container in 8 minutes. we have to find Mass of helium gas diffusing in the same time interval

• molecular mass of SO2 = 64g/mol
• atomic mass of He = 4g/mol

n1 = 4/64 = 1/16 , n2 = m/4

now , (1/16)/(m/4) = √(4/64) = 1/4

⇒m = 1g

hence, mass of helium gas is 1g

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