Question

A solution is to be prepared with a sodium ion concentration of 0.513 mol/L. What mass of sodium sulfate (g) is needed to prepare 1.98 liters of such a solution? Do not enter units with your answer.

Answer 1

We need 72.1 grams of Na2SO4

Step-by-step explanation:

Step 1: Data given

sodium ion concentration = 0.513 mol/L

Volume = 1.98 L

Step 2: The balanced equation

Na2SO4 → 2Na+ + SO4^2-

Step 3: Calculate moles Na+

Moles Na+ = molarity Na+ * volume

Moles Na+ = 0.513 * 1.98 L

Moles Na+ = 1.01574 moles

Step 4: Calculate moles Na2SO4

For 1 mol Na2SO4 we need 2 moles Na+ and 1 mol SO4^2-

For 1.01574 moles Na+ we'll need 1.01574/2 = 0.50787 moles Na2SO4

Step 5: Calculate moles Na2SO4

Mass Na2SO4 = moles Na2SO4 * molar mass Na2SO4

Mass Na2SO4 = 0.50787 moles * 142.04 g/mol

Mass Na2SO4 = 72.1 grams

We need 72.1 grams of Na2SO4

Answer 2

We should weigh 72.1 of sodium sulfate

Step-by-step explanation:

The solution must be made of sodium sulfate. This salt can be dissociated like this:

Na₂SO₄ → 2Na⁺ + SO₄⁻²

From this dissociation we can say that, 2 moles of sodium cation are obtained from 1 mol of salt.

Therefore 0.513 moles of sodium cation will be obtained from 0.2565 moles of salt (0.513 . 1) / 2. The thing is that this number are the moles contained in 1 L of solution, and we need to prepare such 1.98 L

Molarity = mol / Volume (L) → Molarity . volume (L) = mol

0.2565 mol/L . 1.98L = 0.508 moles

These are the moles we should weigh. Let's convert them to moles:

0.508 mol . 142.06 g / 1 mol = 72.1 g