Answer to Question #159330 in Inorganic Chemistry for azai So

Question #159330

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

1
Expert's answer
2021-02-01T03:44:30-0500

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): CM[Ca(C2H3O2)2] = 6.3225 M.

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Comments

JURIN JUNIOR
24.04.21, 05:59

Need your respons as fast as you can..tq so much for your helping..very thankful

Assignment Expert
15.04.21, 12:58

Dear JURIN JUNIOR, You're welcome. We are glad to be helpful. If you liked our service please press like-button beside answer field. Thank you!

JURIN JUNIOR
14.04.21, 20:11

Excellent

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