Dilution - SPM Chemistry

Dilution is a process of adding water to the standard solution lowered the concentration of the solution.
In dilution of solution, we should take note that mole of solute before dilution is equal to the mole of solute after dilution.

Number of mol of solute before dilution = Number of mole of solute after dilution

$\frac{{{M_1}{V_1}}}{{1000}} = \frac{{{M_2}{V_2}}}{{1000}}$
or
${M_1}{V_1} = {M_2}{V_2}$
M₁ = Molarity before dilution
M₂ = Molarity after dilution
V₁ = Volume before dilution
V₂ = Volume after dilution

Example 1
100cm³ of 0.5 mol dm^-3 sodium chloride solution is diluted with distilled water to produce 250 cm³ of solution. Calculate the concentration (in mol dm^-3) of the sodium chloride solution after the dilution.

Answer:
M₁ = 0.5 mol dm^-3
M₂ = ?
V₁ = 100cm³
V₂ = 250 cm³
$\begin{gathered} {M_1}{V_1} = {M_2}{V_2} \hfill \\ (0.5)(100) = {M_2}(250) \hfill \\ {M_2} = \frac{{(0.5)(100)}}{{(250)}} \hfill \\ {M_2} = 0.1mold{m^{ – 3}} \hfill \\ \end{gathered}$

Example 2
Find the volume of 2 mol/dm³ nitric acid that needs to be diluted with distill water to produce 500cm³ of 0.05 mol/dm³ nitric acid.

Answer:
M₁ = 2 mol dm^-3
M₂ = 0.05 mol/dm³
V₁ = ？
V₂ = 500 cm³
$\begin{gathered} {M_1}{V_1} = {M_2}{V_2} \hfill \\ (2){V_1} = (0.05)(500) \hfill \\ {V_1} = \frac{{(0.05)(500)}}{{(2)}} \hfill \\ {V_1} = 12.5c{m^{ – 3}} \hfill \\ \end{gathered}$

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