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This is one of the hw problems that I am strugglingwith. Part a) The hydrocarbon

ID: 76360 • Letter: T

Question

This is one of the hw problems that I am strugglingwith.
Part a) The hydrocarbon naphthalene was frequently used in mothballsuntil recently, when it was discovered that human inhalation ofnaphthalene vapors can lead to hemolytic anemia. Naphthalene is93.71% carbon by mass, and a 0.256 mol sample of naphthalene has amass of 32.8 g. What is the molecular formula ofnaphthalene?
And the answer that I got was C10H8, which turned out to becorrect. Part b)
This compound works as a pesticide in mothballs by sublimation ofthe solid so that it fumigates enclosed spaces with its vaporsaccording to the following equation. If 2.99 g of solid naphthalene isplaced in an enclosed space with a volume of 4.92 L at 25°C, what percentage of thenaphthalene will have sublimed once equilibrium has beenestablished?
And I can't even figure out how to approach part b of thisquestion. Please help. naphthalene(s) naphthalene(g) K = 4.29 10-6 (at 298 K)

Explanation / Answer

b)  Naphthalene (s) Naphthalene(g) ;   K = 4.29 x 10-6                  Molarmass of naphthalene = 128 g/mol         Initial number of moles of naphthalene = 2.99 g /128 g/mol                                                                       = 0.0233 mol               Let x moles of Naphthalene is sublimed                         Naphthalene (s) Naphthalene (g)               I          0.0233mol              0               C                  -x                   x                E         0.0233-x                 x                  K = [ Naphthalene (g)] / [Naphthalene (s)]            4.29 x 10-6 = x / 0.0233-x                 by solving , x = 9.99 x 10-8 mol            percent of naphthalene sublimed at the equilibrium = (9.99 x 10-8 mol / 0.0233 mol) *100                                                                                            = 4.28 x 10-4           
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