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
21 A and B. 21. A solution was made by dissolving 1.63 g MgBr2 in enough...
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