Which region in the IR spectrum could be used to distinguish between butanoic acid and 2-butanone?
1 answer:
Answer:2500-3300
Explanation:We can distinguish between 2-butanone and butanoic acid by analyzing the spectra .
The structure of the two compounds are slightly different as butanoic acid has an OH group whereas for 2-butanone there is no OH group present.
Please refer the attachments for structure of compounds.
Both the compounds will show carbonyl stretching frequency at around 1700-1750cm^{-1}[/tex] due to the presence of carbonyl group in both the compounds.
A broad intense peak at around 3000 will be observed only for butanoic acid due to the presence of OH group and 2-butanone would not show any broad intense peak at around 3000
.
So by analyzing the IR spectra and identifying the intense broad peak of OH we can easily distinguish between butanoic acid and 2-butanone.
Due to hydrogen bonding in between two carboxylic acid molecules ,peak broadening for OH group in butanoic acid takes place.
Generally the ketones show IR stretching in around 1700-1730cm^{-1}[/tex] due to the presence of carbonyl group.
Generally carboxylic acids show IR stretching for carbonyl group at around 1700-1760cm^{-1}[/tex] and a broad intense peak for OH at around 2500-3000cm^{-1}[/tex].
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Answer:
The molarity of urea in this solution is 6.39 M.
Explanation:
Molarity (M) is <em>the number of moles of solute in 1 L of solution</em>; that is
To calculate the molality, we need to know the number of moles of urea and the volume of solution in liters. We assume 100 grams of solution.
Our first step is to calculate the moles of urea in 100 grams of the solution,
using the molar mass a conversion factor. The total moles of 100g of a 37.2 percent by mass solution is
60.06 g/mol ÷ 37.2 g = 0.619 mol
Now we need to calculate the volume of 100 grams of solution, and we use density as a conversion factor.
1.032 g/mL ÷ 100 g = 96.9 mL
This solution contains 0.619 moles of urea in 96.9 mL of solution. To express it in molarity, we need to calculate the moles present in 1000 mL (1 L) of the solution.
0.619 mol/96.9 mL × 1000 mL= 6.39 M
Therefore, the molarity of the solution is 6.39 M.