Question

question 1
0.5 pts
what is the sodium ion concentration (in m) in a 0.355 m sodium arsenide solution?
0.710 m
0.355 m
1.065 m
1.420 m
question 2
0.5 pts
which compound will result in a solution with the highest conductivity?
(nomenclature will never go away.)
1 m scandium sulfate
1 m potassium phosphate
1 m sodium sulfate
2 m chromium(vi) chloride

Explanation:

Question 1

Step1: Determine the formula of sodium arsenide

Sodium arsenide has the formula \( \text{Na}_3\text{As} \) (since sodium is \( \text{Na}^+ \) and arsenide is \( \text{As}^{3-} \), so 3 \( \text{Na}^+ \) balance 1 \( \text{As}^{3-} \)).

Step2: Calculate sodium ion concentration

From the formula, 1 mole of \( \text{Na}_3\text{As} \) dissociates into 3 moles of \( \text{Na}^+ \) ions. So if the concentration of \( \text{Na}_3\text{As} \) is \( 0.355 \, \text{M} \), the concentration of \( \text{Na}^+ \) is \( 3 \times 0.355 \, \text{M} \).

Calculating: \( 3 \times 0.355 = 1.065 \, \text{M} \)

To determine the solution with the highest conductivity, we need to find the compound that dissociates into the most ions (highest total ion concentration) when dissolved. Conductivity is related to the number of ions in solution.

1. 1 M scandium sulfate (\( \text{Sc}_2(\text{SO}_4)_3 \)):

• Dissociation: \( \text{Sc}_2(\text{SO}_4)_3

ightarrow 2\text{Sc}^{3+} + 3\text{SO}_4^{2-} \)

• Total moles of ions per mole of compound: \( 2 + 3 = 5 \)
• Total ion concentration: \( 1 \, \text{M} \times 5 = 5 \, \text{M} \)

1. 1 M potassium phosphate (\( \text{K}_3\text{PO}_4 \)):

• Dissociation: \( \text{K}_3\text{PO}_4

ightarrow 3\text{K}^+ + \text{PO}_4^{3-} \)

• Total moles of ions per mole of compound: \( 3 + 1 = 4 \)
• Total ion concentration: \( 1 \, \text{M} \times 4 = 4 \, \text{M} \)

1. 1 M sodium sulfate (\( \text{Na}_2\text{SO}_4 \)):

• Dissociation: \( \text{Na}_2\text{SO}_4

ightarrow 2\text{Na}^+ + \text{SO}_4^{2-} \)

• Total moles of ions per mole of compound: \( 2 + 1 = 3 \)
• Total ion concentration: \( 1 \, \text{M} \times 3 = 3 \, \text{M} \)

1. **2 M chromium(VI) chloride (\( \text{CrCl}_6 \))? Wait, likely a typo, probably \( \text{CrCl}_3 \) (assuming chromium(VI) might be a typo, but if we take \( \text{CrCl}_3 \)):

• Dissociation: \( \text{CrCl}_3

ightarrow \text{Cr}^{3+} + 3\text{Cl}^- \)

• Total moles of ions per mole of compound: \( 1 + 3 = 4 \)
• Total ion concentration: \( 2 \, \text{M} \times 4 = 8 \, \text{M} \)? Wait, no, if it's 2 M \( \text{CrCl}_3 \), but let's re - check. Wait, maybe the formula is \( \text{Cr(VI)} \) chloride, but chromium(VI) is \( \text{Cr}^{6+} \), so \( \text{CrCl}_6 \) would dissociate as \( \text{CrCl}_6

ightarrow \text{Cr}^{6+}+6\text{Cl}^- \), total ions per mole: \( 1 + 6 = 7 \), total ion concentration: \( 2 \, \text{M} \times 7 = 14 \, \text{M} \)? But this seems odd. Wait, maybe there is a mistake in the problem or my initial assumption. Wait, the original options:

Wait, the options are:

• 1 M scandium sulfate

• 1 M potassium phosphate

• 1 M sodium sulfate

• 2 M chromium(VI) chloride

Wait, let's re - evaluate:

For \( \text{Cr(VI)} \) chloride, the formula is \( \text{CrCl}_6 \) (assuming), dissociation: \( \text{CrCl}_6
ightarrow \text{Cr}^{6+}+6\text{Cl}^- \), so per mole of \( \text{CrCl}_6 \), we get 7 moles of ions. At 2 M, the total ion concentration is \( 2\times7 = 14 \, \text{M} \)

But wait, this is conflicting with the other options. However, maybe the intended formula for chromium(VI) chloride is \( \text{CrO}_2\text{Cl}_2 \) (chromyl chloride), but that is a covalent compound and does not dissociate into ions. So there must be a mistake. Alternatively, maybe it's chromium(III) chloride. Let's assume it's a typo and it's 2 M chromium(III) chloride (\( \text{CrCl}_3 \)):

Dissociation: \( \text{CrCl}_3
ightarrow \text{Cr}^{3+}+3\text{Cl}^- \), total ions per mole: 4, total ion concentration: \( 2\times4 = 8 \, \text{M} \)

But among the given options, if we consider the correct dissociation:

Wait, the first option: 1 M scandium sulfate (\( \text{Sc}_2(\text{SO}_4)_3 \)) dissociates into 2 \( \text{Sc}^{3+} \) and 3 \( \text{SO}_4^{2-} \), total 5 ions per mole, so 5 M total ions.

The fourth option: 2 M chromium(VI) chloride. If we assume the formula is \( \text{CrCl}_6 \) (even though it's unusual), dissociation into \( \text{Cr}^{6+} \) and 6 \( \text{Cl}^- \), total 7 ions per mole. At 2 M, the total ion concentration is \( 2\times7 = 14 \, \text{M} \), which is higher than the other options.

But this is con…

Answer:

1.065 M (corresponding to the option with 1.065 M)

Question 2