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

hey there!

boiling point is the temperature at which the vapor pressure of the liquid equals atmospheric pressure. the vapor pressure of solvent molecules is decreased when a solute is added, so a higher temperature is required to increase the number of solvent molecules in the gas phase above the liquid.

at the freezing point, the vapor pressures of the solid and liquid are equal; a lower temperature is needed to reduce the number of solvent particles above the liquid.

hope this you

tobey

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