Question

How can you dilute solution B containing 20.0 m Cs/mL to give three new solutions with 4 , 3, and $\mathbf{1}\mathbf{\mu gCs}\mathbf{/}\mathbf{mL}$ ?

Short answer

$4\mu gCs/mL$solution can be obtained from the dilution of 50mL solution B in 250mL.

role="math" localid="1663309055250" $3\mu gCs/mL$solution can be obtained from the dilution of 75mL solution B in 500mL .

role="math" localid="1663309044993" $1\mu gCs/mL$solution can be obtained from the dilution of 25mL solution B in 500 mL .

Step-by-step solution

The process for dilution

The process of dilution is used to make low-concentration solutions from higher-concentration solutions.

The dilution is done using the volumetric principle, which is used to calculate the concentration and volume of the diluted solution from a known solution.

${\mathrm{M}}_{\mathrm{con}}{\mathrm{V}}_{\mathrm{con}}={\mathrm{M}}_{\mathrm{dil}}{\mathrm{V}}_{\mathrm{dil}}$

The dilution is done using the volumetric principle, which is used to calculate the concentration and volume of the diluted solution from a known solution.

${\mathrm{M}}_{\mathrm{con}}{\mathrm{V}}_{\mathrm{con}}={\mathrm{M}}_{\mathrm{dil}}{\mathrm{V}}_{\mathrm{dil}}$

Where,

The volume of Concentrated solution is${\mathrm{V}}_{\mathrm{con}}$

The volume of diluted solution ${\mathrm{V}}_{\mathrm{dil}}$

The Concentration of Concentrated solution is ${\mathrm{M}}_{\mathrm{con}}$

The Concentration of diluted solution${\mathrm{M}}_{\mathrm{dil}}$

The process for serial dilution

Serial Dilution - The stepwise dilution method is used to prepare low concentration solutions from higher concentration solutions.

$1\mathrm{\mu gCs}/\mathrm{mL}$solution will be got from the dilution of 25 mL solution B in 500 mL

$4\mathrm{\mu gCs}/\mathrm{mL}$solution will be got from the dilution of 50 mL solution B in 250mL

$1\mathrm{\mu gCs}/\mathrm{mL}$solution will be got from the dilution of 75mL solution B in 500 mL

Create three new solutions

For explanation of the dilution of into 20.0mgCs/mL into 1 , 3 and $4\mathrm{\mu gCs}/\mathrm{mL}$ solutions

Concentration of Solution B is 20.0 mg Cs/mL

The beginning volume of solution B to dilute is determined by plugging in the concentration of solution B and the required final concentrations and volumes in the preceding equation.

$1\mathrm{\mu gCs}/mL$solution will be got from the dilution of 25 mL solution B in 500mL

$$\begin{gathered} {\mathrm{V}}_{\mathrm{con}}=\frac{1\mathrm{\mu gCs}/\mathrm{mL}\times 500\mathrm{mL}}{20\mathrm{mg}/\mathrm{L}} \\ =25\mathrm{mL} \end{gathered}$$

$4\mathrm{\mu gCs}/\mathrm{mL}$ solution will be got from the dilution of solution in

$$\begin{gathered} {\mathrm{V}}_{\mathrm{con}}=\frac{4\mathrm{\mu gCs}/\mathrm{mL}\times 250\mathrm{mL}}{20\mathrm{mg}/\mathrm{L}} \\ =50\mathrm{mL} \end{gathered}$$

$3\mathrm{\mu gCs}/\mathrm{mL}$solution will be got from the dilution of 75 mL solution B in 500mL

$$\begin{gathered} {\mathrm{V}}_{\mathrm{con}}=\frac{3\mathrm{\mu gCs}/\mathrm{mL}\times 250\mathrm{mL}}{20\mathrm{mg}/\mathrm{L}} \\ =75\mathrm{mL} \end{gathered}$$

The process has been explained in the above steps