Calculate the mass, in grams, of lithium fluoride needed to prepare 0.5 L of 0.51 M solution.

Answer 1

Molarity (M) is measured in moles/liter therefore a 0.51 M solution is 0.51 moles/liter.

So 0.51 moles/liter divided by 0.5 liters = 1.02 meaning that this solution contains 1.02 moles of lithium fluoride.

Lithium fluoride is LiF (lithium ion is +1 and fluorine ion is -1) therefore its molar mass (using the periodic table) is 6.941 g/mol (lithium) + 18.998 g/mol (fluorine) = 25.939 g/mol

Since we have 1.02 moles with a molar mass of 25.939 g/mol then (1.02)(25.939) = 26.46 grams of LiF. You would need to round this to the appropriate number of significant figures. I'll let you do that.