Urea, (NH2)2CO, is a product of metabolism of proteins. An aqueous solution is 37.2% urea by mass and has a density of 1.032 g/m
1 answer:
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.
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Answer:
See explanation below
Explanation:
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<u>Answer:</u> 72 grams of water will be produced.
<u>Explanation:</u>
To calculate the number of moles, we use the formula:
....(1)
Mass of propane = 44 grams
Molar mass of propane = 44 grams
Putting values in above equation, we get:
For the reaction of combustion reaction of propane, the equation follows:
By Stoichiometry of the reaction,
1 mole of propane produces 4 moles of water.
So, 1 mole of propane will produce = of water.
Now, to calculate the amount of water, we use equation 1, we get:
Molar mass of water = 18 g/mol
Mass of water produced = 72 grams
Hence, 72 grams of water will be produced.