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

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

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Concentration of Na2CO3 = 0.600 M.

Volume of Na2CO3 = 30.0 ml = 0.030 L.

Polarity = dissolved moles / volume of solution

Now,

moles of Na2CO3 = M * V = 0.600 * 0.030 = 0.018 moles

Na2CO3 ↔ 2Na ^ + + CO3 ^ 2-

By stoichiometry:

1 mole of Na2CO3 produces 2 moles of Na + ion

Thus, 0.018 moles of Na2CO3 would dissolve: 2 * 0.018 = 0.036 moles of Na + ion.

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 + iain