All naturally occuring polysaccharide have one terminal residue a free anomeric carbon why do these polysaccharide not give a po
1 answer:
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Answer:
17,932.69 g/mol is the molecular weight of the substance.
Explanation:
Using Beer-Lambert's law :
Formula used :
where,
A = absorbance of solution = 1.04
c = concentration of solution =?
l = length of the cell = 1 cm
= molar absorptivity of this solution = 18,650
Now put all the given values in the above formula, we get the molar absorptivity of this solution.
c =
V = Volume of the solution in L
Molecular weight of the substance = x
V = 100 mL = 0.1 L
Mass of the substance = 100 mg = 0.1 g
x = 17,932.69 g/mol
17,932.69 g/mol is the molecular weight of the substance.
Answer:
1. Yes
2.The solubility of X is 34.55g/L
Explanation:
Solubility of solute refers to how readily a solute will dissolve in a solvent at a particular temperature. Its the amount of moles or grams required to saturate 1dm or 1 Litre of water.
From the problem, when the liquid was drained off and amount of X which didn't dissolve was measured, it weighed 0.008kg, this means out of 0.027kg, 0.027-0.008 actually dissolved
= 0.019kg*1000 = 19g.
if 19g is required to saturate 550mL at 30°C,
then will saturate 1L
= 34.545g will saturate 1Litre
The solubility thus is 34.55g/L