How many milliliters of 0.748 M Mg(NO3)2 contain 1.700 g of Mg(NO3)2

Answer 1

mL x (1 L/1000 mL) x M x molar mass = grams or you can do it in pieces.
mols Mg(NO3)2 = grams/molar mass
Then M = mols/L. You know M and mols, solve for L and convert to mL.