Question

A sodium hydroxide solution containing \[40\% \] by weight of pure \[{\text{NaOH}}\] has a specific gravity of 1.5. What volume of this solution will be required in the preparation of \[{\text{500 mL}}\] of a \[{\text{0}}{\text{.45M NaOH}}\] solution.

Answer 1

Hint:When a solution is diluted by addition of water, you can use the following formula to calculate the volume needed
\[{M_1} \times {V_1} = {M_2} \times {V_2}\]
Here, \[{M_1},{V_1}\] represents the initial molar concentration and initial volume. Also \[{M_2},{V_2}\] represent the final molar concentration and final volume. Here, initial refers to concentrated stock solution and final refers to diluted solution.

Complete answer:
The weight percent of sodium hydroxide solution is \[40\% \] , Hence, one hundred grams of solution contains forty grams of sodium hydroxide.
The aqueous sodium hydroxide solution has density of 1.5. Here density is obtained from specific gravity.
Divide mass of solution with its density to obtain the volume of the solution
\[
  {\text{volume of the solution = }}\dfrac{{100{\text{ g}}}}{{1.5}} \\
\Rightarrow {\text{volume of the solution}} = 66.66{\text{ ml}}
 \]
 Convert the unit of volume from milliliters to liters by multiplying with 1000.
\[
  {\text{volume of the solution}} = 66.66{\text{ ml }} \times 1000{\text{ mL/L}} \\
\Rightarrow {\text{volume of the solution}} = 0.06666{\text{ L}}
 \]
The molecular weight of sodium hydroxide is \[40{\text{ g/mol}}\]
Divide the weight of sodium hydroxide with its molecular weight to obtain the number of moles.
\[\Rightarrow \dfrac{{40{\text{ g}}}}{{40{\text{ g/mol}}}} = 1{\text{ mol}}\]
Divide the number of moles of sodium hydroxide with the volume of solution (in liter) to obtain the molar concentration of sodium hydroxide solution.
\[\Rightarrow \dfrac{1}{{0.06666}} = 15{\text{M}}\]
Now use the following formula to calculate the volume needed
\[
  {{\text{M}}_{\text{1}}}{\rm{ \times }}{{\text{V}}_{\text{1}}}{\text{ = }}{{\text{M}}_{\text{2}}}{\rm{ \times }}{{\text{V}}_{\text{2}}} \\
\Rightarrow {\text{15M \times }}{{\text{V}}_{\text{1}}}{\text{ = 0}}{\text{.45M \times 500ml}} \\
\Rightarrow {{\text{V}}_{\text{1}}}{\text{ = 15 mL}} \]

Hence, \[{\text{15 mL}}\] of the solution will be needed.

Note:
Molarity is the number of moles of solute present in one litre of solution. Similar concept to molarity is molality. It represents the number of moles of solute present in one kilogram of solvent.