Question

question 6 (1 point)
how many moles of mg⁺² in 6.00 ml of a 6.0 n mg₃(po₄)₂ solution ?
first we have to convert moles into equivalents
mg₃(po₄)₂ → 3 mg⁺² + 2 po₄⁻³
1 mol = 1 equivalents
1 mol mp = 2 mol mg⁺²
3 eq 1 mol
(6.00 x 10⁻³ l)(---------) = 4 eq mp (-------------) = 5 mol mp
1 l 6 eq
7 mol mg⁺²
(8 mol mp)(-----------------) = 9 mol mg⁺²
10 mol mp
a. 250. b. 100.0 c. 2 d. 30.00 e. 6 f. 24
g. 3 h. 36 i. 18.0 j. 1 k. 6.00 x 10⁻³ l. 3.60 x 10⁻²
m. 6.00 x 10⁻² n. 5 o. 3.0 x 10⁻² p. 1.80 x 10⁻²
q. 7.65 x 10⁻¹⁰ r. 2.63 s. 7.89 t. 5.26 u. 95.21
v. 10¹² w. 10⁹ x. 10⁶ y. 10³ z. 105.21

Explanation:

Step1: Calculate equivalents of \(Mg_3(PO_4)_2\)

The normality formula is \(N=\frac{\text{equivalents}}{\text{volume (L)}}\). Given \(N = 6.0\) and \(V=6.00\times10^{-3}\text{ L}\), equivalents \(=N\times V\).
\(\text{equivalents of }Mg_3(PO_4)_2=6.0\times6.00\times 10^{-3}=3.6\times10^{-2}\) (so \(4 = 3.6\times10^{-2}\))

Step2: Convert equivalents to moles of \(Mg_3(PO_4)_2\)

From the dissociation \(Mg_3(PO_4)_2
ightarrow3Mg^{ + 2}+2PO_4^{-3}\), the valency factor (number of equivalents per mole) for \(Mg_3(PO_4)_2\) is \(6\) (because \(3\) \(Mg^{2 +}\) ions, each with a charge of \(+ 2\)). Using the formula \(\text{moles}=\frac{\text{equivalents}}{\text{equivalents per mole}}\)
\(\text{moles of }Mg_3(PO_4)_2=\frac{3.6\times 10^{-2}}{6}=6.0\times10^{-3}\) (so \(5 = 6.0\times10^{-3}\))

Step3: Calculate moles of \(Mg^{2+}\)

Since \(1\) mole of \(Mg_3(PO_4)_2\) gives \(3\) moles of \(Mg^{2+}\)
\(\text{moles of }Mg^{2+}=3\times\text{moles of }Mg_3(PO_4)_2\)
\(\text{moles of }Mg^{2+}=3\times6.0\times10^{-3}=1.8\times10^{-2}\) (so \(9 = 1.8\times10^{-2}\))

Now filling the blanks:

1. \(1\) mole of \(Mg_3(PO_4)_2\) gives \(3\) \(Mg^{2+}\) ions. The valency of \(Mg^{2+}\) is \(2\). The number of equivalents per mole of \(Mg_3(PO_4)_2\) (equivalents per mole \(=\sum(\text{charge}\times\text{number of ions})\)) \(=3\times2 = 6\), so \(1\) \(J\)
2. From \(Mg_3(PO_4)_2

ightarrow3Mg^{2+}+2PO_4^{-3}\), \(1\) \(G\)

1. Normality is \(6.0\), so \(3\) \(D\)
2. As calculated above \(4\) \(L\)
3. As calculated above \(5\) \(K\)
4. Valency factor \(6\) (from step 2) \(6\) \(E\)
5. \(3\) (from dissociation) \(7\) \(G\)
6. Moles of \(Mg_3(PO_4)_2\) \(6.0\times10^{-3}\) \(8\) \(K\)
7. As calculated above \(9\) \(P\)
8. \(1\) (mole ratio denominator) \(10\) \(J\)

Answer:

1. J
2. G
3. D
4. L
5. K
6. E
7. G
8. K
9. P
10. J