1. How many equivalents of Mg2+ are present in a solution that contains 2.50 mol

ID: 952658 • Letter: 1

Question

1. How many equivalents of Mg2+ are present in a solution that contains 2.50 moles of Mg2+?

2. Classify the following electrolytes and write a balanced equation for each of the following:

3. MgCl2(s), CH3OH(l), HF(l)

4. How many grams of NaCl must be combined with water to prepare 500.0 ml of a 0.90% (m/v) physiological saline solution?

5. Calculate the milliliters of a 2.50 M K2SO4 solution to obtain 1.20 moles of K2SO4

6. How would you prepare 250 ml of a 0.225 % (m/v) NaCl solution from a 0.90% (m/v) NaCl stock solution?

(1) A solution with 2.50 moles of Mg2+ contains ( 2 x 2.50) = 5 equivalents of Mg2+

(2) MgCl2 when dissolves in water acts as a strong electrolyte. HF is a weak electrolyte. CH3OH is a non-electrolyte.

(3) MgCl2 -----> Mg2+ + 2 Cl-

HF -----> H+ + F-

(4) Mass of NaCl = ( 500 mL x 0.009 g/ mL ) = 4.5 g

(5) Molarity of K2SO4 solution = 2.50 M

Moles = 1.20

Volume = ( Moles / Molarity) = 0.48 L = 480 mL

(6) Let the required volume of the stock solution be V mL

0.9% x V = 250mL x 0.225 %

V = (250 mL x 0.225 % ) / ( 0.9 % )

V = 62.5 mL

We have to take 62.5 mL of the stock solution and then add distilled water upto 250 mL.

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