Question

How many moles of $N{a}^{+}$ ions are present in $20$ mL of $0.40$ M $N{a}_{3}P{O}_4$?

• A. $0.0080$
• B. $0.024$
• C. $0.050$
• D. $0.20$

Answer 1

Answer: (B)

$0.024$

$1$ molecule of $N{a}_{3}P{O}_4$ dissociates to give $3$ sodium ions.
The number of moles of $N{a}^{+}$ ions present in $20$ mL of $0.40M$ $N{a}_{3}P{O}_4$ $=\frac{20}{1000}\times 0.40\times 3=0.024$.
Thus, $0.024$ mol of $N{a}^{+}$ ions are present in $20$ mL of $0.40M$ $N{a}_{3}P{O}_4$.