Question

Part B: A chemist prepares a solution by adding 415 mg of CoCl2 (MW = 129.84 g/mol ) to a volumetric flask, and then adding water until the total volume of the contents of the flask reaches the calibration line that indicates 100 mL. Determine the molarity of the prepared solution. (moles per litre, 3 sig figs)

Part C: Determine the mass of chloride (MW = 35.45 g/mol ) in grams present in 100 mL of a 0.166 M solution of aqueous FeCl3 (iron(III) chloride) (3 sig figs)

Part D: A beaker contains 447 mL of a 5.5 M HCl(aq) solution. Determine the new concentration of the solution after it is diluted by adding 151 mL of water to the beaker. (3 sig figs)

Answer 1

Part B.) Part B.) moles of Colla mans(q) = 0.415 = 3.2x10 ond! molan man 129.84. malonit - moles of solute molarity = moles - 3.2x10 - 0.0320 M I volume of solution (L) 0,100 Par c.) moles of fell = 0.16680,100 = 0.0166 mol moles of cư = 3x moles offech = 3X0.0166 = 0.0498 moles mans of a = 0.0498 x 35.45 = 1076 g Part D.) ginen, M,=5.5 V = 447 me M2 = 9 V2 = 447+151 = 598 me we know, Miv, = M₂ V2 M, S MINI 5.58 447 V2 598 a 4.11 M

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