Question

Question 14 of 38 > Attempt 1 A 5.29 g mixture contains both lithium fluoride, LiF, and potassium fluoride, KF. If the mixture contains 3.47 g fluorine, what is the mass of the KF in the mixture? mass of KF: 1.82

Answer 1

Answer -

Given,

Mass of the mixture of KF and LiF = 5.29 g

Mass of Fluorine in Mixture = 3.47 g

Mass of KF in Mixture = ?

Let us assume that mass of KF = x grams

So, Mass of LiF in mixture =  5.29 g - x grams

Now, atomic mass of -

F = 19 u

Li = 7 u

K = 39 u

KF = 58 g /mol

LiF = 26 g/mol

Also,

Moles = Mass / Molar Mass

So, Moles of KF in mixture = (x /58) mol

Moles of LiF in mixture = [(5.29 - x)/26] mol

We know that,

1 mole of KF contains 1 mole of F and also, 1 mole of LiF contains 1 mole of F.

Moles of F in KF = (x /58) mol

Moles of F in LiF = [(5.29 - x)/26] mol

Total moles of Fluorine = (x /58) mol + [(5.29 - x)/26] mol

Also,

Mass = Moles * molar Mass

mass of Fluorine in mixture = {(x /58) mol + [(5.29 - x)/26] mol } * 19 u

Put the value of Mass of F (given) and solve,

3.47 g = {(x /58) mol + [(5.29 - x)/26] mol } * 19 u

x = 0.98 g

So. mass of KF in mixture = 0.98 g [Answer]