Home
Class 11
CHEMISTRY
One mole of nitrogen gas at 0.8 atm take...

One mole of nitrogen gas at 0.8 atm takes 38 s to diffuse through a pinhole, whereas one mole of an unknown compound of xenon with fluorine at 1.6 atm takes 57s to diffuse through the same hole. Calculate the molecular formula of the compound.

Text Solution

Verified by Experts

The available data is : `P_(N_(2)) = 0.8` atm, `t_(N_(2)) = 38 s, P_(g) = 1.6` atm , `t_(g) = 57 s`
Rate of diffusion `(r) prop (P)/(sqrt(M))`
`:. (r_(N_(2)))/(r_(g)) = (P_(N_(2)))/(P_(g)) xx sqrt((M_(g))/(M_(N_(2))))`
Since rate of diffusion `(r) prop 1/t`.
`:. (t_(g))/(t_(N_(2))) = (P_(N_(2)))/(P_(g)) xx sqrt((M_(g))/(M_(N_(2))))`
On substituting the value, `57/58 = (0.8)/(1.6) xx sqrt((M_(g))/(28))`
or `sqrt((M_(g))/(28)) = (57)/(38) xx 2 = 3`
Squaring both sides, `(M_(g))/(28) = 9` or `M_(g) = 28 xx 9 = 252`
Let the molecular formula of compound of `Xe` be `Xe F_(n)`.
Molecular mass of `XeF_(n) = 132 + n xx 19`
Comparing (i) and (ii)
`131 + n xx 19 = 252` or `n = (252-131)/(19) = 121/19 = 6.3` or `6`
`:.` Molecular formula `= XeF_(6)`.
Promotional Banner

Similar Questions

Explore conceptually related problems

One mole of nitrogen gas at 0.8 atm takes 38 second of diffuse through a pin hole whereas one mole of an unknown compound of Xenon with fluorine at 1.6 atm takes 56.26 second to diffuse through at same hole, then the molecular formula of the compound is (Atomic mass of Xenon : 131.3 u)

One mole of nitrogen gas at 0.8atm takes 38s to diffuse through a pinhole, while 1 mol of an unknown fluoride of xenon at 1.6 atm takes 57 s to diffuse through the same hole. Calculate the molecular formation of the compound.

One mole of nitrogen gas at 0.8atm takes 38s to diffuse through a pinhole, while 1 mol of an unknown fluoride of xenon at 1.6 atm takes 57 s to diffuse through the same hole. Calculate the value of n in the compound if formula is XeF_n .

One mole of He gas at 0.6 atm takes 40 s to diffuse through a pin hole. Calculate the time required for diffusion of half mole of CH_(4) gas at 1.2 atm through the same hole.

Under identical conditions of pressure and temperature, 4 L of gaseous moxture ( H_(2) and CH_(4) ) effuses through a hole in 5 min whereas 4 L of a gas X of molecular mass 36 takes to 10 min to effuse through the same hole. The mole ratio of H_(2):CH_(4) in the mixture is:

1.40L of an unknown gas requires 57 second to diffuse and the same volume of N_(2) gas takes 84 second to diffuse at the same temperature and pressure. What is the molecular mass of the unknown gas?

A 5L cylinder contained 10 moles of oxygen gas at 27^(@)C . Due to sudden leakage through the hole, all the gas escaped into the atmosphere and the cylinder got empty. If the atmospheric pressure is 1.0atm , calculate the work done by the gas.

20 dm^3 of an unknown gas diffuse through a porous partition in 60 s, whereas 14.1 dm^3 of O_2 under similar conditions diffuse in 30 s. What is the molecular mass of the gas ?

A mixture of N_(2) and a gas 'X' at 300K, is allowed to diffused into empty container of 5.0 L volume. The pressure inside the vessel recorded as 5.5 atm. If 0.9 moles of N_(2) is present in the mixture then calculate molecular mass of 'X'

At 27^(@) C , hydrogen is leaked through a tiny hole into a vessel for 20 min . Another unknown gas at the same T and P as that of H_(2) , is leaked through the same hole for 20 min. After the effusion of the gases the mixture exerts a pressure of 6 atm. The hydrogen content of the mixture is 0.7 mole. If the volume of the container is 3 litre, what is molecular weight of unknown gas ? ("Use" : R = 0.821 "L atm K"^(-1) "mole"^(-1))