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denis23 [38]
2 years ago
12

In your reactions, you diluted the provided stock concentration of ADH for your alcohol. Calculate the actual ADH concentration

in mg/mL used in each of your reactions. (Did this amount vary between reactions?) Show your work. Refer back to Worksheet 1 if you have questions.
Chemistry
1 answer:
Fittoniya [83]2 years ago
5 0

Answer:

10^-3 M.

Explanation:

The chemical compound known as the ADH is a hormone with chemical formula of C46H65N15O12S2, and the

Molar mass of 1084.24 g mol^−1. The chemical compound that is ADH is very important in human body as it helps in regulation of blood and also the regulation of water in human body.

STEP ONE: Convert the value of mg/ml to g/l. That is;

35mg/ml = 35g/l.

STEP TWO: Determine the number of moles.

The number of moles = 35 /35000 = 10^-3 moles.

STEP THREE: Determine the concentration.

Number of moles = concentration × volume.

10^-3 = concentration × 1.

The concentration of ADH = 10^-3 M.

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A student placed 10.5 g of glucose (C6H12O6) in a volumetric fla. heggsk, added enough water to dissolve the glucose by swirling
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<u>Answer:</u> The mass of glucose in final solution is 0.420 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}        .........(1)

Initial mass of glucose = 10.5 g

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

\text{Initial molarity of glucose}=\frac{10.5\times 1000}{180.16\times 100}\\\\\text{Initial molarity of glucose}=0.583M

To calculate the molarity of the diluted solution, we use the equation:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of the concentrated glucose solution

M_2\text{ and }V_2 are the molarity and volume of diluted glucose solution

We are given:

M_1=0.583M\\V_1=20.0mL\\M_2=?M\\V_2=0.5L=500mL

Putting values in above equation, we get:

0.583\times 20=M_2\times 500\\\\M_2=\frac{0.583\times 20}{500}=0.0233M

Now, calculating the mass of final glucose solution by using equation 1:

Final molarity of glucose solution = 0.0233 M

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

0.0233=\frac{\text{Mass of glucose in final solution}\times 1000}{180.16\times 100}\\\\\text{Mass of glucose in final solution}=\frac{0.0233\times 180.16\times 100}{1000}=0.420g

Hence, the mass of glucose in final solution is 0.420 grams

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