Answer to Question #159330 in Inorganic Chemistry for azai So

Question

a) 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

Solution:

Ca(C₂H₃O₂)₂ - calcium acetate

m[Ca(C₂H₃O₂)₂] = 200.0 g

(i):

Molecular mass calculation:

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

The molecular mass of 1 mole Ca(C₂H₃O₂)₂ is 158.166 g mol^-1.

(ii):

One mol of Ca(C₂H₃O₂)₂ has a mass of 158.166 grams.

Thus 200 grams of 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

1.2645 mol of Ca(C₂H₃O₂)₂ has a mass of 200.0 grams.

(iii):

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

Hence,

Number molecules = (1.2645 mol) × (6.022×10²³ molecules / 1 mol) = 7.615×10²³ molecules

Number molecules of Ca(C₂H₃O₂)₂ is 7.615×10²³ molecules.

(iv):

At STP, one 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

1.2645 mol of Ca(C₂H₃O₂)₂ occupies a volume equal to 28.325 L.

(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 grams (according to the (ii) calculation).

Solution volume = 200 mL = 0.2L

Hence,

Molarity of Ca(C₂H₃O₂)₂ = 1.2645 mol / 0.2 L = 6.3225 mol/L = 6.3225 M

The molarity of a solution is 6.3225 M.

Answer:

(i): M[Ca(C₂H₃O₂)₂] = 158.166 g mol^-1;

(ii): n[Ca(C₂H₃O₂)₂] = 1.2645 mol;

(iii): N[Ca(C₂H₃O₂)₂] = 7.615×10²³ molecules;

(iv): V[Ca(C₂H₃O₂)₂] = 28.325 L;

(v): C_M[Ca(C₂H₃O₂)₂] = 6.3225 M.