Chemistry, asked by medicaltopper2022, 6 months ago

What is the molarity of Na+ion in a
solution prepared by dissolving
0.550 gram of Na2SO4 in a volume
of 100 ml water.

Answers

Answered by kundanconcepts800
15

Answer:

molarity = no.of mol of solute/volume of solution in litres

= 0.550/142.04/0.1L

= 0.038 M

[Na+] = 0.076M

Answered by dualadmire
1

The molarity of the Na+ ion in the solution is 0.076 M.

Given: Mass of Na2SO4 = 0.550 g,

           The volume of water = 100 ml

To Find: The molarity of Na+ ion in the solution.

Solution:

  • We know that the molarity of a solution can be calculated using the formula,

               Molarity (M) = number of moles of solute (n) / Volume ( in L )

  • We know that the molarity of a compound is equal to the molarity of its constituent ions in the ratio of the molecular formula.

 Hence, 1 mole of Na2SO4 contains= 2 moles of Na+ ion = 1 mole of S atom

Coming to the numerical, we are given that,

        Mass of Na2SO4 = 0.550 g,

        Molecular mass of Na2SO4 = ( 2×23 + 32 + 4×16 ) = 142 g

Thus, the number of moles of Na2SO4 = Mass of Na2SO4 /Molecular mass

                                                           = 0.55 / 142

                                                           = 3.873 × 10^-3

        The volume of water = 100/1000 L

                                     = 0.1 L

Hence, the molarity of Na2SO4 = number of moles of Na2SO4 /Volume

                                                     =  3.873 × 10^-3 / 0.1  M

                                                     =  0.038 M

Now, as said earlier,

1 mole of Na2SO4 contains= 2 moles of Na+ ion

the molarity of Na2SO4 = 0.038 M

Hence, the molarity of Na+ ions = 2 × molarity of Na2SO4

                                                     = 2 × 0.038 M

                                                     = 0.076 M

Hence, the molarity of the Na+ ion in the solution is 0.076 M.

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