How many atoms are there in 2.43 g of calcium?
1 answer:
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Answer:
0.03g/mL
Explanation:
Given parameters include:
Five μL of a 10-to-1 dilution of a sample; This implies the Volume of dilute sample is given as 5 μL
Dilution factor = 10-to-1
The absorbance at 595 nm was 0.78
Mass of the diluted sample = 0.015 mg
We need to first determine the concentration of the diluted sample which is required in calculating the protein concentration of the original solution.
So, to determine the concentration of the diluted sample, we have:
concentration of diluted sample =
= (where ∝ was use in place of μ in the expressed fraction)
= 0.003 mg/μL
The dilution of the sample is from 10-to-1 indicating that the original concentration is ten times higher; as such the protein concentration of the original solution can be calculated as:
protein concentration of the original solution = 10 × concentration of the diluted sample.
= 10 × 0.003 mg/μL
= 0.03 mg/μL
= 0.03g/mL
Hence, the protein concentration of the original solution is known to be 0.03g/mL
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.