Explain why the solutions of SbCl3(aq) in our lab also contain HCl. Why would adding a few drops of this solution to 300 mL of w
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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.
Aromatic compound has continuous cyclic structure with( 4n+2)π electrons (Huckels rule), where n = 0,1,2…
Here number of pi electron are 6, where 4 from two double bond and 2 from nitrogen non-bonding electrons, hence it has total 6 pi electrons, therefore
6= ( 4n+2)π
4 = 4n
n =1
Hence it is an aromatic compound
