Answer:
The correct answer is option b, that is, 2.1 M Na₃PO₄.
Explanation:
The solution with the largest concentration of ions will possess the highest conductivity.
a) 3.0 M NaCl
NaCl ⇔ Na⁺ + Cl⁻
Here the total number of ions is 2, therefore, the concentration of ions is 3.0 × 2 = 6.0 M
b) 2.1 M Na₃PO₄
Na₃PO₄ ⇔ 3 Na⁺ + PO₄³⁻
Here the total number of ions i 4. Therefore, the concentration of ions is
2.1 × 4 = 8.4 M.
c) 2.4 M CaCl₂
CaCl₂ ⇔ Ca²⁺ + 2Cl⁻
The total number of ions is 3. Therefore, the concentration of ions is
2.4 × 3 = 7.2 M
d) 3.2 M NH₄NO₃
NH₄NO₃ ⇔ NH₄⁺ + NO₃⁻
The total number of ions is 2. The concentration of ions will be,
3.2 × 2 = 6.4 M
Hence, the highest conductivity will be of 2.1 M Na₃PO₄.
The theoretical yield of Ca(OH)₂ : 42.032 g
<h3>Further explanation</h3>
Given
31.8 g of CaO
Required
The theoretical yield of Ca(OH)₂
Solution
Reaction
CaO + H₂O⇒Ca(OH)₂
mol CaO (MW=56 g/mol) :
= mass : MW
= 31.8 g : 56 g/mol
= 0.568
From equation, mol Ca(OH)₂ = mol CaO = 0.568
Mass Ca(OH)₂ (MW=74 g/mol) :
= 0.568 x 74
= 42.032 g
They have six valence electrons