Answer to Question #341426 in Inorganic Chemistry for Amzar

Question

200.0 g of Ca(C₂H₃O₂)₂ was found in the chemistry laboratory. 

      Calculate

          i. the molecular mass of 1 mole Ca(C₂H₃O₂) _2.

         ii. the number of  moles of Ca(C₂H₃O₂)₂  that exists   

        iii. the number of Ca(C₂H₃O₂)₂ molecules present

        iv.  the volume at STP  if Ca(C₂H₃O₂)₂ is converted to gas.

        v.  the molarity of a solution made by dissolving 200.0 g of      

                Ca(C₂H₃O₂)₂ in 200 ml of water

       

Answer 1

I. M[Ca(C₂H₃O₂)₂] = 40.078 + (12.0107 × 2 + 1.00794 × 3 + 15.9994 × 2) × 2 = 158.166 g mol^-1

II. 1 mol Ca(C₂H₃O₂)₂ has a mass of 158.166 g

200 g Ca(C₂H₃O₂)₂ is (200 g Ca(C₂H₃O₂)₂) × (1 mol Ca(C₂H₃O₂)₂ / 158.166 g Ca(C₂H₃O₂)₂) = 1.2645 mol Ca(C₂H₃O₂)₂

III. 1 mole of any substance contains 6.022×10²³ atoms/molecules

N[Ca(C₂H₃O₂)₂] = (1.2645 mol) × (6.022×10²³ molecules / 1 mol) = 7.615×10²³ molecules

IV. At STP, 1 mole of any gas occupies a volume of 22.4 L

thus 1.2645 mol of Ca(C₂H₃O₂)₂ occupies: (1.2645 mol × 22.4 L) / 1 mol = 28.325 L of Ca(C₂H₃O₂)₂

V. Molarity of Ca(C₂H₃O₂)₂ = Moles of Ca(C₂H₃O₂)₂ / Solution volume

1.2645 mol of Ca(C₂H₃O₂)₂ has a mass of 200.0 g (according to the (ii) calculation)

Solution volume = 200 mL = 0.2L, C_M Ca(C₂H₃O₂)₂ = 1.2645 mol / 0.2 L = 6.3225 mol/L = 6.3225 M