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The molecular weight of silver bromide (AgBr) is 187.77 g/mole. The presence of the ions in solution can be shown as- AgBr (insoluble) ⇄ +
.
45.00 mL of the aliquot contains 0.6485 g of AgBr. Thus 1000 mL of the aliquot contains ×1000 = 14.411 gm-mole. Thus the solubility product
of AgBr = [
]×
.
Or, 5.0× =
(the given value of solubility product of AgBr is 5.0×
and the charge of the both ions are same).
Thus S = (5.00×)
= 7.071×
g/mL.
Thus the concentration of Br or HBr is 7.071×
g/mL.
Explanation:
a) Using Beer-Lambert's law :
Formula used :
where,
A = absorbance of solution = 0.945
c = concentration of solution = ?
l = length of the cell = 1.20 cm
= molar absorptivity of this solution =
()
14.16 μM is the molarity of the red dye solution at the optimal wavelength 519nm and absorbance value 0.945.
b)
1 L of solution contains moles of red dye.
Mass of moles of red dye:
c) In order to dilute red dye solution by 5 times, we will need to add 1 L of water to solution of given concentration.
Concentration of red dye solution =
Concentration of red solution after dilution = c'
The final concentration of the diluted solution is