Question

21 A and B.

21. A solution was made by dissolving 1.63 g MgBr2 in enough water to give 50.0 mL of solution. Then, 10.0 mL of this solution were transferred to another flask and diluted to a volume of 75.0 mL. a. What is the molarity of this final solution? va: 75.0ML V(= 50.0mL 1.639 mg By i mols 1 184.119 mg 1 X 24. 31 Br: 2x 179.904 159.808 184.118 :8.858 103 mois TRIM OSL b. How many moles of bromide ions are contained in this final solution? 1.639 mg Br Imol /2 mol Bri 184.11g / limol mg Br2) :.01 mol Br

Answer 1

21)

Molar mass of MgBr2,

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

= 1*24.31 + 2*79.9

= 184.11 g/mol

mass(MgBr2)= 1.63 g

use:

number of mol of MgBr2,

n = mass of MgBr2/molar mass of MgBr2

=(1.63 g)/(1.841*10^2 g/mol)

= 8.853*10^-3 mol

volume , V = 50 mL

= 5*10^-2 L

use:

Molarity,

M = number of mol / volume in L

= 8.853*10^-3/5*10^-2

= 0.1771 M

use dilution formula

M1*V1 = M2*V2

1---> is for stock solution

2---> is for diluted solution

Given:

M1 = 0.1771 M

V1 = 10 mL

V2 = 75 mL

use:

M1*V1 = M2*V2

M2 = (M1*V1)/V2

M2 = (0.1771*10)/75

M2 = 0.0236 M

Answer: 0.0236 M

b)

Mol of MgBr2 = M(MgBr2)*Volume

= 0.0236 M * 0.075 L

= 1.77*10^-3 mol

1 mol of MgBr2 has 2 mol of Br-

So,

Mol of Br- = 2*1.77*10^-3 mol

= 3.54*10^-3 mol

Answer: 3.54*10^-3 mol