Question

Calculate the number of Na ions and Cl ions and total number of ions in 14.5g of NaCl.​

Answer 1

To find the number of Na and Cl ions in 14.5g of NaCl, 14.5g is first converted to moles (0.072 mol), then multiplied by Avogadro's number to obtain approximately 4.34 x 10^22 ions for each type. The total number of ions is thus about 8.68 x 10^22.

Step-by-step explanation:

To calculate the number of Na ions and Cl ions in 14.5g of NaCl, we follow these steps:

Convert from grams to moles of NaCl using the molar mass of NaCl in the unit conversion factor. Given that the formula mass of NaCl (sodium chloride) is 58.44 amu, which is the sum of the atomic masses of one sodium atom (22.99 amu) and one chlorine atom (35.45 amu), we find out that 14.5 g of NaCl is equivalent to 0.072 mol NaCl.

Determine the number of moles of ions present in the solution. Since NaCl dissociates into two ions (Na+ and Cl-) for every one mole of NaCl, we use the conversion 2 mol ions/1 mol NaCl. The result from the conversion is 0.072 mol NaCl x 2 mol ions/1 mol NaCl = 0.14 mol ions.

Therefore, there are 0.072 mol of Na ions and 0.072 mol of Cl ions because of the 1:1 ratio in NaCl. To find the total number of each type of ion, we use Avogadro's number (6.022 x 1023 ions/mol). Multiplying the number of moles of Na and Cl ions by Avogadro's number gives us approximately 4.34 x 1022 ions for each Na and Cl. The total number of ions is double this amount, which is roughly 8.68 x 1022 ions.

Answer 2

Number of Na ions in 14.5 g of NaCl is 1.49 × 10²³.

Number of Cl ions in 14.5 g of NaCl is 1.49 × 10²³.

Total number of ions = 1.49 × 10²³ + 1.49 × 10²³ = 2.98 × 10²³.

Step-by-step explanation:

1 mole of any compound contains 6.023 × 10²³ molecules.

molecular weight of NaCl is 23 + 35.5 = 58.5 g.

so, 58.5 grams of NaCl makes 1 mole

⇒ 14.5 g of NaCl =
`(14.5)/(58.5)` = 0.248 moles.

⇒ 0.248 mole contains 0.248 × 6.023×10²³ molecules

= 1.49 × 10²³ molecules.

And 1 molecule contains 1 Na ion and 1 Cl ion.

⇒ number of Na ions in 14.5 g of NaCl is 1.49 × 10²³.

⇒ number of Cl ions in 14.5 g of NaCl is 1.49 × 10²³.

Total number of ions = 1.49 × 10²³ + 1.49 × 10²³ = 2.98 × 10²³.