Question

How many grams of lithium fluoride is required to make 1.2 L of a 3.5 M solution?

Answer 1

Answer:

109.2g

Explanation:

First, let us obtain the number of mole of lithium fluoride in the solution. This is illustrated below:

Molarity of lithium fluoride (LiF) = 3.5M

Volume = 1.2L

Number of mole of LiF =?

Molarity = mole/Volume

Mole = Molarity x Volume

Number of mole LiF = 3.5 x 1.2

Number of mole LiF = 4.2 moles

Now let us convert 4.2 moles to grams in order to obtain the desired result. This illustrated below:

Molar Mass of LiF = 7 + 19 = 26g/mol

Number of mole of LiF = 4.2 moles

Mass of LiF =?

Mass = number of mole x molar Mass

Mass of LiF = 4.2 x 26

Mass of LiF = 109.2g

Therefore, 109.2g of lithium fluoride is required.