Answer to Question #97189 in General Chemistry for Archie

Question #97189
A 5.97 g mixture contains both lithium fluoride, LiF, and potassium fluoride, KF. If the mixture contains 3.90 g fluorine, what is the mass of the KF in the mixture?
1
Expert's answer
2019-10-23T07:01:09-0400

Q97189


Solution:


Given;

Total weight of mixture "(LiF+KF)=5.97gms."

Let, mass of "KF" in the mixture be "x gms."

"\\implies" mass of "LiF" in mixture "=(5.97-x)gms."


Note:

Atomic weight of "F=19gms."

Atomic weight of "Li =7gms."

Atomic weight of "K = 39 gms."


"moles=mass\/(molecular weight)"

Thus, moles of "KF=x\/58"

and moles of "LiF = (5.97-x)\/26"


1 mol of KF contains 1 mol of F atoms. Similarly, 1 mol of LiF contains 1 mol of F atoms.

Thus,

moles of F in "KF=moles of KF=x\/58" ---(1)

moles of F in "LiF =moles of LiF= (5.97-x)\/26" ---(2)


From (1) & (2), we get;

Total moles of Fluorine "=(x\/58)+((5.97-x)\/26)"

Hence, total weight of Fluorine in sample "= moles*Atomic weight"

"=((x\/58)+((5.97-x)\/26))*19gms."

"=3.90 gms." ---(given)

Now, solving the equation for x, we get;

"26x +(5.97*58)-58x=3.9*58*26\/19"

"22x=346.26-309.537"

"\\therefore x=36.723\/22"

"=1.669gms." (Answer)

Thus, mass of KF in the mixture is "1.669gms."



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