If you had a 0.5 M KCl solution, how much solute would you have in moles, and what would the solute be?

Since KCl has a molecular weight of 74.5, how many grams/L would this be?

37.25 grams/L.

Explanation:

Molarity (M) is defined as the no. of moles of solute dissolved per 1.0 L of the solution.

M = (no. of moles of KCl) / (volume of the solution (L))

∵ no. of moles of KCl = (mass of KCl) / (molar mass of KCl)

∴ M = [ (mass of KCl) / (molar mass of KCl) ] / (volume of the solution (L))

∴ (mass of KCl) / (volume of the solution (L)) = (M) * (molar mass of KCl) = (0.5 M) * (74.5 g/mol) = 37.25 g/L.

So, the grams/L of KCl = 37.25 grams/L.