Question

1. You need to make an aqueous solution of 0.241 M zinc bromide for an experiment...

1. You need to make an aqueous solution of 0.241 M zinc bromide for an experiment in lab, using a 300 mL volumetric flask. How much solid zinc bromide should you add?

2.How many milliliters of an aqueous solution of 0.132 M calcium chloride is needed to obtain 8.00 grams of the salt?

1. In the laboratory you dissolve 22.3 g of silver nitrate in a volumetric flask and add water to a total volume of 250 mL.

What is the molarity of the solution?  M.

What is the concentration of the silver cation?  M.

What is the concentration of the nitrate anion?  M.

2. In the laboratory you dissolve 15.6 g of iron(III) fluoride in a volumetric flask and add water to a total volume of 125 mL.

What is the molarity of the solution?  M.

What is the concentration of the iron(III) cation?  M.

What is the concentration of the fluoride anion?  M.

Homework Answers

Answer #1

1)

volume , V = 300 mL

= 0.300 L

use:

number of mol,

n = Molarity * Volume

= 0.241*0.3

= 7.23*10^-2 mol

Molar mass of ZnBr2,

MM = 1*MM(Zn) + 2*MM(Br)

= 1*65.38 + 2*79.9

= 225.18 g/mol

use:

mass of ZnBr2,

m = number of mol * molar mass

= 7.23*10^-2 mol * 2.252*10^2 g/mol

= 16.28 g

Answer: 16.3 g

2)

Molar mass of CaCl2,

MM = 1*MM(Ca) + 2*MM(Cl)

= 1*40.08 + 2*35.45

= 110.98 g/mol

mass(CaCl2)= 8.00 g

use:

number of mol of CaCl2,

n = mass of CaCl2/molar mass of CaCl2

=(8 g)/(1.11*10^2 g/mol)

= 7.209*10^-2 mol

use:

M = number of mol / volume in L

0.132 = 7.209*10^-2/ volume in L

volume = 0.546 L

volume = 546 mL

Answer: 546 mL

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