Question
A 1/5 mmol (yes, millimol... no, it isn't a typo_ sample of XeFn h as a mass of 0.311 g. What is the value of n?
The answer is 4, but why?
The answer is 4, but why?
Answers
Kayla
omg that should say 1.50 mmol
DrRebel
Given 0.311g sample = 1.50 mmole sample = 1.5x10^-4 mole sample
Sample mole weight = (0.311)g/(1.5x10^-4)mol = 207.33 g/mol
fwt(Xe) + n·fwt(F) = 207.33 g/mol
(131.3)g/mol + n·(19.0)g/mol = 207.33g/mol
(n·19)g/mol = (207.33 - 131.3)g/mol
n = [(207.33 - 131.3)g/mol]/(19)g/mol
n = 74.03/19 = 4
XeF4
Sample mole weight = (0.311)g/(1.5x10^-4)mol = 207.33 g/mol
fwt(Xe) + n·fwt(F) = 207.33 g/mol
(131.3)g/mol + n·(19.0)g/mol = 207.33g/mol
(n·19)g/mol = (207.33 - 131.3)g/mol
n = [(207.33 - 131.3)g/mol]/(19)g/mol
n = 74.03/19 = 4
XeF4
Kayla
thank you :D
DrRebel
You are most welcome...
DrReb(at)g.m.a.i.l.c.o.m
DrReb(at)g.m.a.i.l.c.o.m