How many sodium ions are present in aqueous sodium phosphate if the sample size is 500 milliliters and the concentration is 2 Molarity?
To determine the number of sodium ions present in aqueous sodium phosphate, we need to use the formula:
Number of ions = (Concentration in Molarity) x (Volume in liters) x (Number of ions per molecule)
In this case, the concentration is 2 Molarity and the volume is 500 milliliters, which is equal to 0.5 liters. The sodium phosphate formula is Na3PO4, so one molecule of sodium phosphate contains 3 sodium ions (Na+).
Using the formula, we can calculate the number of sodium ions:
Number of ions = (2 Molarity) x (0.5 liters) x (3 ions/molecule)
= 1 x 0.5 x 3
= 1.5 moles of Na+ ions
To determine the actual number of sodium ions, we need to multiply the number of moles by Avogadro's number (6.022 × 10^23).
Number of ions = (1.5 moles) x (6.022 × 10^23 ions/mole)
≈ 9.033 × 10^23 sodium ions
Therefore, there are approximately 9.033 × 10^23 sodium ions present in the given aqueous sodium phosphate solution.