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:

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.

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