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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