Question

1. How many grams of zinc chloride are needed to make 245 mL of a 0.769 M solution? 2. Calculate the concentration of the ions present in a 3.0 M sodium phosphate solution. in coolad hon 20 3. Calculate the concentration of a solution prepared by dissolving 35.0 g of aluminum carbonate in enough water to prepare 500.0 mL of solution. 4. The A student diluted 45.7 mL of a 3.78 M aluminum carbonate solution to a final volume of 380.0 mL. What is the new molarity? 5. How many grams of lead (II) nitrate are needed to react with 118.0 mL of 1.11 M H2SO.? 6. It takes 22.4 mL of 0.234 M HCl to titrate 35.8 mL of a NaOH solution. What is the molarity of the base? 7. 325 mL of 0.375 M barium chloride were mixed with 50.0 mL of silver nitrate 0.501 M. a. What will be the maximum number of grams of solid product that can be obtained? b. Which is the reactant in excess and by how many ml?

Answer 1

Date Volume = 245 me Let . molarity Molarity mass Oo 769 M of zucle be Number of moles of solute volume of solution in litres Number of moles- gluen Mass Molecular Mass x 136 Molecular mass of zulla - 136 g/mol So molarity С 136 Xlooo 245 To convert volume into litres 0.7 69=x Xlooo 136 X 245 0.769 x136x245 1000 2 5 623.08 2 - 1000 = 25.6 2 of ions present Calculate the concentration in 3.0 M 28 odium Phosptate solution Naz Poy lag) I mole of NazPou gives 3 moles of Nat & Imole of I mole of a + 3 Nat + Poy Every

Date Iso concentration of wat ions = 34 [Nagpoy = 34 3 19.0m lx (Naz poy] 1 X 3.0 30M Concentration of poy a 3 Volume Soome Given Mass of Aluminum Carbonates = 35.0 Molecular Mass of Alumunlu in Carbonate C Alg (CO₂) ] 233.9 g/mol Couceud ration (lublarity) = number of mot moles of solu Given mass of solute volume of solution an lit Molecular Mass ofs solete Yolume of solution in litres 35 233.9 Q Xloo Solo 35 X 2 = 0.299 M 233.9 M. v. = Mg Vg 3.78 x 45.7 = Mg x 380.0 M2 = 3.78 X45.7 380 0.45 M