Question

eq32: do the tube dilution, final dilution, and solve for v1 for each tube. tube dilution 1/1 final dilution 1/1 write a reply edit view insert format tools table

Explanation:

Step1: Recall dilution formula

The dilution formula is $C_1V_1 = C_2V_2$.

Step2: For the first - tube (no dilution)

Since the tube dilution is $1/1$ and final dilution is $1/1$, and assuming we want to find the volume $V_1$ to get a certain concentration in the same - concentration solution. Let's say we consider the initial concentration $C_1 = 20\ g/mL$ and we want to find the volume to get the same concentration in the first tube. If we take a volume $V_1$ from the stock and put it into the first tube, and since there is no dilution ($C_1 = C_2$), for any amount we take from the stock, the relationship holds. Let's assume we want to make up a final volume $V_2$ of $10\ mL$. Using $C_1V_1 = C_2V_2$ with $C_1 = C_2 = 20\ g/mL$ and $V_2=10\ mL$, we get $V_1 = 10\ mL$.

Step3: For the second tube

We know that the concentration changes from $C_1 = 20\ g/mL$ to $C_2 = 10\ g/mL$. Using the dilution formula $C_1V_1 = C_2V_2$. Let the final volume $V_2 = 10\ mL$. Then $V_1=\frac{C_2V_2}{C_1}=\frac{10\ g/mL\times10\ mL}{20\ g/mL}=5\ mL$.

Step4: For the third tube

The initial concentration $C_1 = 20\ g/mL$ and final concentration $C_2 = 2\ g/mL$, and $V_2 = 10\ mL$. Using $C_1V_1 = C_2V_2$, we have $V_1=\frac{C_2V_2}{C_1}=\frac{2\ g/mL\times10\ mL}{20\ g/mL}=1\ mL$.

Step5: For the fourth tube

The initial concentration $C_1 = 20\ g/mL$ and final concentration $C_2 = 1\ g/mL$, and $V_2 = 10\ mL$. Using $C_1V_1 = C_2V_2$, we get $V_1=\frac{C_2V_2}{C_1}=\frac{1\ g/mL\times10\ mL}{20\ g/mL}=0.5\ mL$.

Step6: For the fifth tube

The initial concentration $C_1 = 20\ g/mL$ and final concentration $C_2 = 0.1\ g/mL$, and $V_2 = 10\ mL$. Using $C_1V_1 = C_2V_2$, we obtain $V_1=\frac{C_2V_2}{C_1}=\frac{0.1\ g/mL\times10\ mL}{20\ g/mL}=0.05\ mL$.

Answer:

For the first tube: $V_1 = 10\ mL$; For the second tube: $V_1 = 5\ mL$; For the third tube: $V_1 = 1\ mL$; For the fourth tube: $V_1 = 0.5\ mL$; For the fifth tube: $V_1 = 0.05\ mL$