Question

for a 25.0 % m/v mgcl₂ solution.

1. what is the molarity ? the gfw for magnesium chloride is 95.21 g. (hint: find

the g in 1000 ml, then convert to moles)
there are 25.0 g in 100. ml, so there are 250. g in 1000. ml

1. g 1 g 2 mol

-------- = (--------)(----------) = 3 m

1. ml 1 l 4 g

1. what is the osmolarity ?

mgcl₂ → mg⁺² + 2 cl⁻¹ 1 mol = 5 osmol
6 mol 7 osmol
(--------)(------------) = 8 osmol/l
9 l 10 mol
a. 250. b. 100.0 c. 2 d. 30.00 e. 6 f. 24 g. 3 h. 36

i. 18.0 j. 1 k. 6.00×10⁻³ l. 3.60×10⁻² m. 6.00×10⁻² n. 5

o. 3.0×10⁻² p. 1.80×10⁻² q. 7.65×10⁻¹⁰ r. 2.63 s. 7.89

t. 5.26 u. 95.21 v. 10¹² w. 10⁹ x. 10⁶ y. 10³

Explanation:

Step1: Calculate moles of \(MgCl_2\)

Molarity formula is \(M=\frac{n}{V}\) (where \(n\) is moles and \(V\) is volume in liters). Given \(25.0\%\ m/v\) means \(25.0\ g\) in \(100\ mL\). So in \(1000\ mL\) (which is \(1\ L\)), mass of \(MgCl_2\) is \(250\ g\). Moles \(n=\frac{m}{M}\), where \(m = 250\ g\) and molar mass \(M=95.21\ g/mol\).

Step2: Substitute values into molarity formula

\(M=\frac{250\ g}{1\ L\times95.21\ g/mol}\)

Answer:

1. A. \(250\)
2. G. \(3\)
3. R. \(2.63\) (since \(\frac{250}{95.21}\approx2.63\))
4. U. \(95.21\)
5. G. \(3\) (because \(MgCl_2\) dissociates into \(Mg^{2 +}+2Cl^{-}\), so \(1\ mol\ MgCl_2 = 3\ osmol\))
6. R. \(2.63\) (molarity from part 1)
7. G. \(3\)
8. R. \(7.89\) (\(2.63\times3 = 7.89\))
9. J. \(1\)
10. J. \(1\)