Question

eq 31: do the tube dilution, final dilution, and solve for v1 for each tube.
20 ml
20 g/ml of x1
2 g/ml
0.2 g/ml
0.02 g/ml
0.002 g/ml
tube dilution 1/1
final dilution 1/1

Explanation:

Step1: Recall dilution formula

$C_1V_1 = C_2V_2$

Step2: For the first - tube (since initial and final concentrations and volumes are the same as per given 1/1 dilutions)

Let $C_1 = 20\ g/mL$, $V_1$ is unknown, $C_2 = 20\ g/mL$, $V_2=20\ mL$. Substituting into $C_1V_1 = C_2V_2$, we get $20V_1=20\times20$. Solving for $V_1$, we divide both sides by 20: $V_1=\frac{20\times20}{20}=20\ mL$.

Step3: For the second tube

Let $C_1 = 20\ g/mL$, $V_1$ is unknown, $C_2 = 2\ g/mL$, $V_2 = 20\ mL$. Using $C_1V_1 = C_2V_2$, we have $20V_1=2\times20$. Then $V_1=\frac{2\times20}{20}=2\ mL$.

Step4: For the third tube

Let $C_1 = 20\ g/mL$, $V_1$ is unknown, $C_2 = 0.2\ g/mL$, $V_2 = 20\ mL$. From $C_1V_1 = C_2V_2$, we get $20V_1=0.2\times20$. So $V_1=\frac{0.2\times20}{20}=0.2\ mL$.

Step5: For the fourth tube

Let $C_1 = 20\ g/mL$, $V_1$ is unknown, $C_2 = 0.02\ g/mL$, $V_2 = 20\ mL$. Using the dilution formula $C_1V_1 = C_2V_2$, we have $20V_1=0.02\times20$. Then $V_1=\frac{0.02\times20}{20}=0.02\ mL$.

Step6: For the fifth tube

Let $C_1 = 20\ g/mL$, $V_1$ is unknown, $C_2 = 0.002\ g/mL$, $V_2 = 20\ mL$. From $C_1V_1 = C_2V_2$, we get $20V_1=0.002\times20$. So $V_1=\frac{0.002\times20}{20}=0.002\ mL$.

Answer:

For the first tube: $V_1 = 20\ mL$
For the second tube: $V_1 = 2\ mL$
For the third tube: $V_1 = 0.2\ mL$
For the fourth tube: $V_1 = 0.02\ mL$
For the fifth tube: $V_1 = 0.002\ mL$