Answer:
A. [tex]C_2=1.5\frac{mg}{mL}[/tex]
B. [tex]C_2=0.075\frac{mg}{mL}[/tex]
C. [tex]C_2=0.01\frac{mg}{mL}[/tex]
D. [tex]C_2=0.001\frac{mg}{mL}[/tex]
Explanation:
Hello.
In this case, we must compute the final concentration in all the cases so we solve for it in the given equation:
[tex]C_2=\frac{C_1V_1}{V_2}[/tex]
Thus, we proceed as follows:
A. Here, the final diluted solution includes the 300 μL of the 5 mg/ml-BSA and the 700 μL of TBS.
[tex]C_2=\frac{300\mu L*5\frac{mg}{mL} }{(300+700)\mu L}\\\\C_2=1.5\frac{mg}{mL}[/tex]
B. Here, the final diluted solution includes the 50 μL of the 1.5 mg/ml-BSA, the 450 μL of water and the 500 μL of TBS.
[tex]C_2=\frac{50\mu L*1.5\frac{mg}{mL} }{(50+450+500)\mu L}\\\\C_2=0.075\frac{mg}{mL}[/tex]
C. Here, the final diluted solution includes the 10 μL of the 1 mg/ml-BSA and the 990 μL of TBS.
[tex]C_2=\frac{10\mu L*1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.01\frac{mg}{mL}[/tex]
D. Here, the final diluted solution includes the 10 μL of the 0.1 mg/ml-BSA and the 990 μL of TBS.
[tex]C_2=\frac{10\mu L*0.1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.001\frac{mg}{mL}[/tex]
Best regards.
