Concentration of Na2CO3 = 0.600 M
Volume of Na2CO3 = 30.0 ml = 0.030 L
Molarity = moles of solute/volume of solution
Now,
moles of Na2CO3 = M * V = 0.600*0.030 = 0.018 moles
Na2CO3 ↔ 2Na^+ + CO3^2-
As per stoichiometry:
1 mole of Na2CO3 produces 2 moles of Na+ ions
Therefore, 0.018 moles of Na2CO3 would dissociate to give: 2*0.018 = 0.036 moles of Na+ ions.
Now,
1 mole of Na+ has 6.023 * 10^23 ions
therefore, 0.036 moles of Na+ would correspond to:
= 0.036 * 6.023 * 10^ 23 = 2.17 *10^22 Na+ ions
