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

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.

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.