Step 8: Solution Calculations (8) Dre unit analysis, clearly show the set-ups fo

Question

Step 8: Solution Calculations (8) Dre unit analysis, clearly show the set-ups for your calculations Using u wish to prepare 50.0 mL of 2.30 x 10-2 M solution of your drug. 1. Calculation. (hint-how many grams will you need of your drug?) Yo 2. Describe how to prepare this solution from the pure drug and deionized water Analytical balances and volumetric flasks are drug dissolves in water, at least to the extent necessary for this problem.) Use complete sentences in your description of the solution preparation process. available to you. (Assume that your 3. A patient is given 2.0mL of a 3.0% (w/v) solution of your drug, how many milligrams of the drug did they receive? 4. What is the molarity of a 3.00% (w/v) solution of glucose, C6H12O6?

Explanation / Answer

Ans. #1. Moles of C8H15N7O2S3 required = Molarity x Volume of solution in liters

                                                            = (2.30 x 10-2 M) x 0.050 L

                                                            = 0.00115 mol

Required mass of drug = Required moles x Molar mass

                                                = 0.00115 mol x (337.45108 g/ mol)

                                                = 0.3881 g

#2. The volume of solution in not mentioned. Let the required volume = 1.0 L

Preparation of drug solution:

Step 1: Accurately weigh and transfer 0.3881 g powdered drug into a 1.- L grade A standard volumetric flask with small amount of water.

Step 2: Dissolve the drug by gently swirling the flask.

Step 3: Once dissolved, make the final volume upto the mark.

Step 4: Put the stopper, and mix the solution by inverting several times to get s homogenate solution.

It sis the desired solution, 2.30 x 10-2 M drug.

#3. A 3.0 % w/v means there is 3.0 gram of solute (here, drug) in 100.0 mL of solution.

Amount of drug in 2.0 mL sample = Volume of sample x [Drug]

                                                            = 2.0 mL x (3.0 g/ 100.0 mL)

                                                            = 0.060 g

                                                            = 60.0 mg

#4. For a 3.0 % w/v glucose solution-

Mass of glucose = 3.0 g

Volume of solution = 100.0 mL = 0.100 L

Now,

            Moles of glucose = 3.0 g / (180.0 g) = 0.016667 mol

And,

            Molarity of glucose = Moles of glucose / Volume of solution in liters

                                                = 0.016667 mol / 0.100 L

                                                = 0.16667 M

                                                = 1.667 x 10-1 M

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.