Answer:
6,666.66 micro liter of water and protein stock we will need to add to obtain the target concentration and volume.
Step-by-step explanation:
Concentration of given solution
![C_1 = 0.15 mg/mL](https://img.qammunity.org/2020/formulas/chemistry/college/m407cq7718tjwvk9v50vrdkkr7zazupo1e.png)
1 mL = 1000 μL , 1 mg = 1000 μg
![C_1=0.15 mg/mL=(0.15* 1000 \mu g)/(1* 1000 \mu L)=0.15 \mu g/\mu L](https://img.qammunity.org/2020/formulas/chemistry/college/p7zyjjfy3binleil1fi3bf5b9kws3taxsr.png)
The volume of the given solution =
![V_1= V](https://img.qammunity.org/2020/formulas/chemistry/college/qxfddfwmi01jseahy8q0u6s84u69254jq8.png)
Concentration of required solution =
![C_2=10 \mu g/\mu L](https://img.qammunity.org/2020/formulas/chemistry/college/boggywxdrxvcqz8sz0jm7lrp6i5hid7d47.png)
Volume of required solution =
![V_2=100 \mu L](https://img.qammunity.org/2020/formulas/chemistry/college/1zvw9qm3xz4bcq8jqvppx7wdreash4pxdp.png)
![C_1V_1=C_2V_2](https://img.qammunity.org/2020/formulas/mathematics/college/617jrrlkpl4y3ise68ncr406u2dyik09gs.png)
![V=(C_2V_2)/(C_1)=(10 \mu g/\mu L* 100 \mu L)/(0.15 \mu g/\mu L)=6,666.66 \mu L](https://img.qammunity.org/2020/formulas/chemistry/college/yzxva3mmp6jhjtq3po1usagmmlkf5i6hx1.png)
6,666.66 micro liter of water and protein stock we will need to add to obtain the target concentration and volume.