Find the normality of the solution containing 5.300 g/l of na2co3 ?

   Normality = 0.1 N

Explanation:

Data given:

weight of Na₂CO₃ = 5.3g/L

Normality: ?

Normality:

It is the expression of concentration for a solution.

Normality is the gram equivalent of a compound in the per liter of solution.

Equivalent:

one equivalent can react with hydrogen ion or replace hydrogen ion.

Formula for Normality:

  Normality = gram equivalent of solute / volume of solution in L (1)

we know

   gram equivalent of solute = weight/ equivalent weight (2)

So by combining equation 1 and 2 the formula will be

             N = weight / equivalent weight / volume of solution in L (3)

Now first we have to find equivalent weight.

As in Na₂CO₃ there are 2 Na that can replace hydrogen ion in a solution,

So, we have 2 equivalent

The Molar mass of the Na2CO3 = (23x2 + 12 + 16x3)

                                      Na2CO3 = 106 g/mol

Now to find equivalent weight:

                  equivalent weight:  molar mass / no. of equivalent (n)

                  equivalent weight:  106 g/mol / 2

                  equivalent weight:  53 g/mol

Put the values in Equation 2:

               gram equivalent = weight/ eq. weight

               gram equivalent = 5.3 / 53

               gram equivalent = 0.1

Now put this value of gram equivalent in equation 1

       Normality = gram equivalent of solute / volume of solution in L

       Normality = 0.1 /1L

      Normality = 0.1 N

So the Normality of the solution is 0.1N

answer:

the answer is 6 valence electrons.

explanation:

valence is the # of electrons on the outer most shell of an atom.

oxygen has 6.

although, the valency of oxygen is 2, because it only needs to make 2 covalent bonds, or gain 2 electrons, to achieve the stable octet configuration of a noble gas.

that's it!

if you need further assistance, let me know: )

~ trueboss72