A6.00l tank at −6.4°c is filled with 8.62g of boron trifluoride gas and 16.9g of chlorine pentafluoride gas. you can assume both gases behave as ideal gases under these conditions. calculate the mole fraction of each gas. round each of your answers to 3 significant digits.

mole fraction boron trifluoride :  0.496

mole franction chlorine pentafluoride gas : 0.504

Explanation:

This is a problem where we have to remember a definition to solve it right away and where there is some additional data which can make us lose sight of the easy solution.

Mole fraction of component A is defined as the ratio of the moles A over the total number of moles in the mixture:

Χₐ = nₐ/ ( na + nb + nc +)

For a two gas mixture it reduces to:

Χₐ = na/(na + nb)

we have the masses so we can convert them to moles as follows:

na = ma/MWₐ      

MW boron triflouride = 67.82 g/mol

nₐ = 8.62 g / 67.82 g/mol  = 0.127 mol

MW chlorine pentafluoride : 130.45 g/moL

nb = 16.9 g / 130.45 g/mol = 0.129 mol  

( MW chlorine pentafluoride : 130.45 g/moL)

Χₐ = 0.127 mol  / (0.127+0.129) mol = 0.496

Χb = 0.129 mol /  (0.127+0.129) mol =0.504   ( also since its 2 gases we can calculate the mole fraction of b by using the expression Xa + Xb =1)

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8