please help me with honors chem molarity

Question

Calculate each of the following quantities.

(a) grams of solute in 164.8 mL of 0.207 M calcium acetate

  ---------------g

(b) molarity of 584 mL solution containing 21.1 g of potassium iodide

 ----------------------M

(c) moles of solute in 145.6 L of 0.812 M sodium cyanide

----------------------------- mol

Answer 1

molar mass calcium acetate, Ca(C2H3O2)2 = 158.2 g/mol

164.8 ml x 1 L / 1000 ml x 0.207 mol / L x 158.2 g / mol = 5.40 g calcium acetate (3 sig. figs.)

molar mass potassium iodide, KI = 166 g/mol

584 ml x 1 L/1000 ml = 0.584 L

21.1 g KI x 1 mol / 166 g = 0.127 mols

molarity = 0.127 mols / 0.584 L = 0.218 M

145.6 L x 1 L x 0.812 mol / L = 118.2 mols