Question

how many osmoles are there in 6.00 ml of a 6.0 n mg₃(po₄)₂ solution ?
first we have to convert moles into osmoles
mg₃(po₄)₂ → 3 mg⁺² + 2 po₄⁻³
1 mol = 6 eq
1 mol = 1 osmoles
2 eq 3 mol
(6.00 x 10⁻³ l)(----------) = 4 eq (----------) = 5 mol mp
1 l 6 eq
7 osmol
(8 mol mp)(-------------) = 9 osmol
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

We know that normality formula is \(N=\frac{\text{equivalents}}{\text{volume (L)}}\). Given \(N = 6.0\) and \(V=6.00\times10^{- 3}\text{ L}\).
Equivalents \(=\text{Normality}\times\text{Volume}\)
\(eq=(6.0\ \text{eq/L})\times(6.00\times10^{-3}\text{ L})\)
\(eq = 3.6\times10^{-2}\)

Step2: Convert equivalents to moles

Since \(1\ \text{mol}=6\ \text{eq}\) (from \(Mg_3(PO_4)_2
ightarrow3Mg^{2 +}+2PO_4^{3-}\), total \(3\times2 + 2\times3=12\) charges, but normality is based on total ion charges. For \(Mg_3(PO_4)_2\), \(n = 6\) (sum of absolute values of cation and anion charges per formula unit). So \(\text{mol}=\frac{\text{eq}}{n}\)
\(\text{mol}=\frac{3.6\times10^{-2}\text{ eq}}{6\ \text{eq/mol}}\)
\(\text{mol}=6.0\times10^{-3}\)

Step3: Convert moles to osmoles

From \(Mg_3(PO_4)_2
ightarrow3Mg^{2+}+2PO_4^{3 -}\), \(1\ \text{mol}\) of \(Mg_3(PO_4)_2\) gives \(3 + 2=5\) moles of ions. But in terms of osmoles (based on total number of particles in solution), for \(Mg_3(PO_4)_2\), \(1\ \text{mol}\) of \(Mg_3(PO_4)_2\) gives \(5\) osmoles.
\(\text{osmol}=\text{mol}\times5\)
\(\text{osmol}=(6.0\times10^{-3}\text{ mol})\times5\)
\(\text{osmol}=3.0\times10^{-2}\)

Answer:

1. \(5\)
2. \(6.0\)
3. \(6.0\times10^{-3}\)
4. \(3.6\times10^{-2}\)
5. \(6.0\times10^{-3}\)
6. \(6\)
7. \(5\)
8. \(6.0\times10^{-3}\)
9. \(3.0\times10^{-2}\)
10. \(1\)