Question

How many total moles of ions are released when each of the following samples dissolves in water? (a) 0.49 mol of NH4Cl WebAssign will check your answer for the correct number of significant figures. 0.98 Correct: Your answer is correct. mol (b) 25.3 g of Ba(OH)2 · 8 H2O WebAssign will check your answer for the correct number of significant figures. 0.240 Correct: Your answer is correct. mol (c) 1.86 1020 formula units of LiCl WebAssign will check your answer for the correct number of significant figures. mol

Answer 1

The total moles of ions released can be calculated by considering the formula of the compound and the number of ions it produces in solution.

Step-by-step explanation:

To calculate the total moles of ions released when a sample dissolves in water, we need to consider the formula of the compound and the number of ions it produces in solution.

(a) 0.49 mol of NH4Cl: NH4Cl dissociates into NH4+ and Cl- ions, so the total number of moles of ions released is 0.49 mol x 2 = 0.98 mol.

(b) 25.3 g of Ba(OH)2 · 8 H2O: Ba(OH)2 dissociates into Ba2+ and OH- ions, so the total number of moles of ions released is 25.3 g / 315.46 g/mol x 3 = 0.240 mol.

(c) 1.86 x 1020 formula units of LiCl: Each formula unit of LiCl dissociates into Li+ and Cl- ions, so the total number of moles of ions released is 1.86 x 1020 formula units x 2 = 3.72 x 1020 mol.

Answer 2

(a) 0.98 mol

(b) 0.241 mol

(c) 6.18 × 10⁻⁴ mol

Step-by-step explanation:

(a)

Let´s consider the ionization of NH₄Cl.

NH₄Cl(s) → NH₄⁺(aq) + Cl⁻(aq)

1 mole of NH₄Cl produces 2 moles of ions. The moles of ions produced by 0.49 moles of NH₄Cl are:

0.49 mol NH₄Cl × (2 mol ions/1 mol NH₄Cl) = 0.98 mol ions

(b)

Let's consider the ionization of Ba(OH)₂· 8 H₂O.

Ba(OH)₂· 8 H₂O(s) → Ba²⁺(aq) + 2 OH⁻(aq) + 8 H₂O(l)

The molar mass of Ba(OH)₂· 8 H₂O is 315.48 g/mol. The moles of Ba(OH)₂· 8 H₂O corresponding to 25.3 g are:

25.3 g × (1 mol/315.48 g) = 0.0802 mol

1 mole of Ba(OH)₂· 8 H₂O produces 3 moles of ions. The moles of ions produced by 0.0802 moles of Ba(OH)₂· 8 H₂O are:

0.0802 mol Ba(OH)₂· 8 H₂O × (3 mol ions/ 1 mol Ba(OH)₂· 8 H₂O) = 0.241 mol ions

(c)

Let's consider the ionization of LiCl.

LiCl(s) → Li⁺(aq) + Cl⁻(aq)

1 mole of LiCl contains 6.02 × 10²³ formula units of LiCl. The moles of LiCl that contain 1.86 × 10²⁰ formula units are:

1.86 × 10²⁰ fu × (1 mol LiCl/6.02 × 10²³ fu) = 3.09 × 10⁻⁴ mol LiCl

1 mole of LiCl produces 2 moles of ions. The moles of ions produced by 3.09 × 10⁻⁴ moles of LiCl are:

3.09 × 10⁻⁴ mol LiCl × (2 mol ions/1 mol LiCl) = 6.18 × 10⁻⁴ mol ions