Question

1. prepare 45 ml of 0.15 m naoh from 0.8 m naoh solution
2. you add 30 ml of saline solution to 3 ml of serum. calculate the dilution factor of the serum?
3. a 1/10 dilution is diluted 1/100 and then again 1/50. what is the final dilution?
4. 1 ml of a 4.0 m nacl solution is diluted 1/10, rediluted 1/5 and rediluted 1/25. what is the final dilution factor? what is the concentration of the final dilution? what is the volume of the final solution (in l)?

Explanation:

Step1: Solve problem 2

Use the dilution formula $M_1V_1 = M_2V_2$. Here, $M_2 = 0.15\ M$, $V_2=45\ ml$, $M_1 = 0.8\ M$. We need to find $V_1$.
$V_1=\frac{M_2V_2}{M_1}=\frac{0.15\ M\times45\ ml}{0.8\ M}=8.44\ ml$. So, take $8.44\ ml$ of the $0.8\ M\ NaOH$ solution and dilute it to a final volume of $45\ ml$ with water.

Step2: Solve problem 3

The dilution factor (DF) is calculated as $\frac{\text{Final volume}}{\text{Volume of sample}}$. The final volume is $30\ ml + 3\ ml=33\ ml$ and the volume of the serum (sample) is $3\ ml$. So, $DF=\frac{33\ ml}{3\ ml}=11$.

Step3: Solve problem 4

When multiple dilutions are done, the final dilution is the product of the individual dilution factors. The first - dilution factor is $\frac{1}{10}$, the second is $\frac{1}{100}$ and the third is $\frac{1}{50}$. So the final dilution is $\frac{1}{10}\times\frac{1}{100}\times\frac{1}{50}=\frac{1}{50000}$.

Step4: Solve problem 5

The individual dilution factors are $\frac{1}{10}$, $\frac{1}{5}$ and $\frac{1}{25}$. The total dilution factor (TDF) is $\frac{1}{10}\times\frac{1}{5}\times\frac{1}{25}=\frac{1}{1250}$.
The initial concentration $C_1 = 4.0\ M$. Using the formula $C_1V_1 = C_2V_2$, and since $V_1 = 1\ ml$ and $TDF=\frac{V_2}{V_1}$, $V_2=1250\ ml = 1.25\ L$. And $C_2=\frac{C_1}{TDF}=\frac{4.0\ M}{1250}=0.0032\ M$.

Answer:

1. Take 8.44 ml of 0.8 M NaOH solution and dilute to 45 ml.
2. 11
3. $\frac{1}{50000}$
4. Dilution factor: $\frac{1}{1250}$, Concentration: 0.0032 M, Volume: 1.25 L