Chemistry, asked by seshaiahtumati85, 7 months ago

One mole of N, gas at 0.8 atr, takes 38 seconds to diffuse through a pin hole where as one mole of an unknown compound of Xenon with fluorine
at twice the pressure of N, takes 55 seconds to diffuse through the same pin hole. How many lone pairs are around xenon in xenon fluande
compound? (Given Atomic masses Xe = 121 y, F = 19 u)​

Answers

Answered by Anonymous
3

Explanation:

your complete question is -->One mole of nitrogen gas at 0.8 atm takes 38 seconds to diffuse through a pin hole whereas one mole of unknown compound of xenon and fluorine at 1.6 atm takes 57 seconds to diffuse through same hole . find the molecular formula of compound.

from Graham's law of diffusion,

e., t2/t1 = (P1/P2)√{M2/M1}

given, t1 = 38sec, t2 = 57sec , M1 = 28g/mol, P1 = 0.8atm, P2 = 1.6atm and M2 = ?

now, 57/38 = (0.8/1.6)√{M2/28}

squaring both sides,

57²/38² = (1/4)(M2/28)

or, M2 = (57² × 28 × 4)/38² = 252

hence, molecular weight of compound is 252 g/mol.

compound contains Xenon and Fluorine.

atomic weight of Xenon = 131

and atomic mass of Fluorine = 19

and general formula of Xenon fluoride is XeFx

so, 131 + 19x = 252

or, 19x = 252 - 131 = 121

or, x = 6.36. ≈ 6

hence, molecular formula is XeF6

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