Question

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?

Answer 1

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 $1.669gms.$