Answer:
We should weigh 72.1 of sodium sulfate
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
Answer:
We need 72.1 grams of Na2SO4
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
