How many ions are present in 30.0 ml of 0.600 m na2co3 solution?

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

it is a solid

explanation: