Q. 9. 109 - An Introduction to General, Organic, and Biological Chemistry

Question

What is the initial volume, in milliliters, needed to prepare each of the following diluted solutions? $\left(9.5\right)$

a. $250$ $mL$ of $3.0$% $\left(m/v\right) HCI$ from $10.0$% $\left(m/v\right) HCI$

b. $500$. $mL$ of $0.90$% $\left(m/v\right) NaCI$ from $5.0$% $\left(m/v\right) NaCI$

c. $350$. $mL$ of $2.00$$M NaOH$ from $6.00$$M NaOH$

Step-by-Step Solution

Answer

Dilute solutions are needed to prepare each of the following solutions:

 the initial volume  and milliliters.

a. $250 mL$of $3.0\% \left(m/v\right) HCI$ from $10.0\% \left(m/v\right) HCI$

b. $500. mL$of $0.90\% \left(m/v\right) NaCI$ from $5.0\% \left(m/v\right) NaCI$

c. $350. mL$of $2.00M NaOH$from $6.00M NaOH$

Step 1: Introduction:

They needed dilute solutions in the original volume and millilitres to make each of the following solutions.

Step 2: Given:

To prepare each of the following solutions, they needed dilute solutions in the initial volume and millilitres.

a. $250 mL$of $3.0\% \left(m/v\right) HCI$ from $10.0\% \left(m/v\right) HCI$

b. $500. mL$of $0.90\% \left(m/v\right) NaCI$ from $5.0\% \left(m/v\right) NaCI$

c. $350. mL$of $200M NaOH$ from $6.00 M NaOH$

Step 3: According to the dilution expression:

According to the dilution expression:

  ${\mathrm{C}}_1\times {\mathrm{V}}_1={\mathrm{C}}_2\times {\mathrm{V}}_2$

Here,

${\mathrm{C}}_1$= concentration of the concentrated solution

 ${\mathrm{V}}_1$= volume of the concentrated

  ${\mathrm{C}}_2$ = concentration of the diluted solution

   ${\mathrm{V}}_2$= volume of the diluted solution

Step 4: (part a) The initial volume of HCI solution:

According to the problem, 

${\mathrm{C}}_2 = 3.0\% \left(\mathrm{m}/\mathrm{v}\right)$

${\mathrm{V}}_2 = 250\mathrm{mL}$

${\mathrm{C}}_1 = 10.0\% \left(\mathrm{m}/\mathrm{v}\right)$

${\mathrm{V}}_{1 = ?}$

Rearrange the dilution expression to solve for the unknown quantity ${\mathrm{V}}_1$

$\begin{array}{r}{V}_1=\frac{C_2\times V_2}{C_1}\end{array}$

$\begin{array}{r}\\ V_1=\frac{3.0\mathrm{\%}(\mathrm{m}/\mathrm{v})\times 250\mathrm{mL}}{10.0\mathrm{\%}(\mathrm{m}/\mathrm{v})}\end{array}$

$= 75mL$

The required initial volume of HCI solution is $75 mL$

Step 5: (part b) The initial volume of NaCI solution:

According to the problem,

${\mathrm{C}}_2 = 0.9\% \left(\mathrm{m}/\mathrm{v}\right)$