How many sodium ions are in 4543.3 g of NaCl?

4.68x10²⁵ ions of Na⁺

Explanation:

First of all, we dissociate the salt:

NaCl(aq) → Na⁺(aq) + Cl⁻(aq)

An aqueous solution of NaCl dissociates in chlorides anions and sodium cations. Ratio is 1:1, per 1 mol of NaCl, we have 1 mol of Na⁺

We determine the moles of salt: 4543.3 g . 1mol / 58.45 g = 77.7 moles

77.7 moles are the amount of NaCl, therefore we have 77.7 moles of Na⁺.

We count the ions:

1 mol fo Na⁺ has 6.02x10²³ ions

77.7 moles of Na⁺ must have (77.7 . 6.02x10²³) / 1 = 4.68x10²⁵ ions of Na⁺

We have 4.68 * 10^25 Na+ ions

Explanation:

Step 1: Data given

Mass of NaCl = 4543.3 grams

Number of Avogadro = 6.02 * 10^23 particles / mol

Molar mass NaCl = 58.44 g/mol

Step 2: The equation

NaCl → Na+ + Cl-

For 1 mol NaCl we have 1 mol Na+ ions and 1 mol Cl- ions

Step 3: Calculate moles NaCl

Moles NaCl = mass NaCl / molar mass NaCl

Moles NaCl = 4543.3 grams / 58.44 g/mol

Moles NaCl = 77.74 moles

Step 4: Calculate moles Na+ ions

For 1 mol NaCl we have 1 mol Na+ ions and 1 mol Cl- ions

For 77.74 moles NaCl we have 77.74 moles Na+

Step 5: Calculate Na+ ions

Number of Na+ ions = moles Na+ * number of Avogadro

Number of Na+ ions = 77.74 moles * 6.02 * 10^23 / moles

Number of Na+ ions = 4.68 * 10^25  Na+ ions

We have 4.68 * 10^25 Na+ ions

16

explanation:

the marks go by 4s