Question

How many grams of sodium sulfate are needed to make 1.22 L of a 2.10 M sodium sulfate solution? Give your answer here with correct sig figs and units (Hint: You'll need to find the formula and molar mass of sodium sulfate)

Answer 1

To make 1.22 L of a 2.10 M solution of sodium sulfate, 363.81 grams of sodium sulfate are needed, calculated using the molar mass of Na2SO4 (142.04 g/mol).

Step-by-step explanation:

To calculate how many grams of sodium sulfate (Na2SO4) are needed to make 1.22 L of a 2.10 M sodium sulfate solution, we first need to determine the molar mass of Na2SO4. The formula for sodium sulfate is Na2SO4, which contains 2 sodium ions (Na), 1 sulfur ion (S), and 4 oxygen ions (O). The molar masses of these elements (approximate to two decimal places) are 22.99 g/mol for Na, 32.07 g/mol for S, and 16.00 g/mol for O. Therefore, the molar mass of Na2SO4 is:

(2 × 22.99 g/mol) + (1 × 32.07 g/mol) + (4 × 16.00 g/mol) = 142.04 g/mol.

Now, we can use the molarity equation, which is:

Molarity (M) = moles of solute/liters of solution

To find the moles of solute required:

2.10 M = moles of Na2SO4 / 1.22 L

moles of Na2SO4 = 2.10 M × 1.22 L = 2.562 moles.

Finally, we convert moles to grams using the molar mass:

2.562 moles × 142.04 g/mol = 363.8088 g.

Therefore, 363.81 grams of sodium sulfate are needed (rounded to four significant figures), matching the given sig figs of the molarity and volume.