Question

You are given a protein solution with a concentration of 0.15 mg/ml.

Answer 1

Explanation:

The given data is as follows.

         Initial concentration = 0.15 mg/ml,        

        Final concentration = 0.1 mg/ml, (as $\frac{0.1 microgram}{1 microliter} \times \frac{10^{-3}}{1 microgram} \times \frac{1 microliter}{10^{-3}ml}$)

        Final volume = $100 microliter \times \frac{10^{-3} ml}{1 microliter}$ = 0.1 ml

According to the dilution formula we get the following.

              $C_{i} \times V_{i} = C_{f} \times V_{f}$

or,                        $V_{i}$ = $\frac{C_{f} \times V_{f}}{C_{i}}$

Putting the given values into the above formula we get the following.

                   $V_{i}$ = $\frac{C_{f} \times V_{f}}{C_{i}}$

                               = $\frac{0.1 mg/ml \times 0.1 ml}{0.15 mg/ml}$

                               = 0.0667 ml

                               = $6.67 \times 10^{-2} ml \times \frac{1 microliter}{10^{-3} ml}$                                

                              = 66.7 microliter

This means that volume of protein stock which is required is 66.7 ml. Hence, calculate the volume of water required as follows.

                  Volume of water required = Total volume - volume of protein stock

                                                     = (100 - 66.7)  microliter

                                                     = 33.3 microliter

Thus, we can conclude that we need 33.3 microliter of water and 66.7 microliter of protein.