Question

1. How do you prepare 500 ml of 30% v/v ethanol?

2. How do you prepare 225 ml of 25% w/v NaCl?

3. You have a 20 ml of a 10% w/v solution of ammonium sulfate. How many grams of ammonium sulfate are in this solution?

4. You want to run a cesium chloride gradient to purify a sample of DNA in an ultracentrifuge. You wish to fill a 15 ml test tube with five equal layers of 5%, 10%, 15%, 20%, and 25% cesium chloride. How much total cesium chloride will you need?

5. The concentration of the hemoglobin monomer (17 kDa) is approximately 15 grams per deciliter of blood. The average human blood volume is 5 L. What is the concentration of the hemoglobin monomer in your body in millimolar?

5. You have a 4 M stock solution of phosphate buffer. How do you prepare 75 ml of 0.8 M phosphate buffer from this stock?

5. You have a 25 ml of protein at 10 mg/ml. If you concentrate this solution to 5 ml, what is the new concentration?

Answer 1

In the first question, v/v means volume/volume i.e volume of solute present per 100ml of solution.

To prepare v/v concentration, first of all, we have to calculate the required solute volume (which is in our case 30% ) to prepare. 100ml solution.

Or we can say, 30 ml solute is mixed in 100-30= 70 ml of water to prepare 30+70=100 ml of 30% ethanol. 70ml of water is to be taken because if we mix more volume, the total volume of the solution will exceed 100ml.

Therefore, to prepare 500ml of 30% ethanol, the calculation is-

500×30/100= 150ml ethanol is required.

And the required volume of water is- 500-150ml= 350ml

Similarly, to prepare w/v we have to calculate the amount of solute is required first to make 100ml of solution. But remember, in w/v the amount of water will be 100ml, not the( 100ml-calculated volume of the solute)water. Because NaCl will dissolve in water and not increase the total volume of the solution.

​Therefore, to prepare 225ml of 25% NaCl solution( 25% means 25 gram in 100ml of water), the calculation will be-

225×25/100= 5625/100= 56.25 gram of NaCl to be mixed with 225ml of water.

In the 3rd question, 10% w/v ammonium sulfate means 10gram of ammonium sulfate is present in 100ml water.

We have 20 ml of that solution.

Therefore, the amount of ammonium sulfate present in 20ml of solution is-

20×10/100 gram

= 200/100 gram

= 2 gram

Hence, 2 gram of ammonium sulfate is present in 20ml solution.