How many liters of a .88 M solution can be made with 25.5 grams of lithium fluoride (#LiF#)?

Question

18576 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-16T15:25:34

#L = 1.10 L# of solution

The Molarity #M# is calculated by the equation comparing moles of solute to liters of solution

#M=(mol)/L#

For this question we are given the Molarity 0.88M

We are told the solute is a 25.2 gram sample of LiF, Lithium Fluoride

We can convert the mass of LiF to moles by dividing by the molar mass of LiF

Li = 6.94
F = 19.0

LiF = 25.94 g/mole

#25.2 cancel(grams) x (1 mol)/(25.94cancel(grams))# = #0.97# moles

Now we can take the the molarity and the moles and calculate the Liters of solution

#M=(mol)/L#

#ML= mol#

#L = (mol)/M#

#L = (0.97mol)/(0.88M)#

#L = 1.10 L# of solution