Question

How many grams of \[{\text{NaOH}}\] are required to prepare $200{\text{ mL}}$ of a $0.450{\text{ M}}$ solution?

Answer 1

Hint: Using the given molarity and volume of solution calculate the number of moles of \[{\text{NaOH}}\]. Determine the molar mass of \[{\text{NaOH}}\]. Using the number of moles and molar mass of \[{\text{NaOH}}\] calculate the mass of \[{\text{NaOH}}\].

Complete solution:
We have to calculate the mass of \[{\text{NaOH}}\] in $200{\text{ mL}}$ of a $0.450{\text{ M}}$ solution.
Using the given molar concentration and volume of the solution calculate the number of moles of \[{\text{NaOH}}\] as follows:
The equation of molarity is as follows:
\[{\text{Molarity}}\left( {\text{M}} \right) = \dfrac{{{\text{Number of moles of solute}}\left( {{\text{mol}}} \right)}}{{{\text{Volume of solution}}\left( {\text{L}} \right)}}\]
Rearrange the equation for the number of moles as follows:
\[{\text{Number of moles of solute}}\left( {{\text{mol}}} \right) = {\text{Molarity}}\left( {\text{M}} \right) \times {\text{Volume of solution}}\left( {\text{L}} \right)\]
Substitute $200{\text{ mL}} = 200 \times {10^{ - 3}}{\text{ L}}$ for the volume of the solution, $0.450{\text{ M}}$ for the molarity. Thus,
\[{\text{Number of moles of NaOH}} = {\text{0}}{\text{.450 M}} \times 200 \times {10^{ - 3}}{\text{ L}}\]
\[{\text{Number of moles of NaOH}} = 90 \times {10^{ - 3}}{\text{ mol}}\]
Thus, the number of moles of \[{\text{NaOH}}\] in $200{\text{ mL}}$ of a $0.450{\text{ M}}$ solution are \[90 \times {10^{ - 3}}{\text{ mol}}\].
Now, we can convert these calculated moles of \[{\text{NaOH}}\] to mass as follows:
The mathematical equation related to moles, mass and molar mass is as follows:
\[{\text{Number of moles of solute}}\left( {{\text{mol}}} \right) = \dfrac{{{\text{Mass}}\left( {\text{g}} \right)}}{{{\text{Molar mass}}\left( {{\text{g/mol}}} \right)}}\]
We can rearrange this equation for mass as follows:
\[{\text{Mass}}\left( {\text{g}} \right) = {\text{Number of moles of solute}}\left( {{\text{mol}}} \right) \times {\text{Molar mass}}\left( {{\text{g/mol}}} \right)\]
Now, substitute \[90 \times {10^{ - 3}}{\text{ mol}}\] for moles of \[{\text{NaOH}}\] , \[40.0{\text{ g/mol}}\] for molar mass of \[{\text{NaOH}}\]. Thus,
\[{\text{Mass of NaOH}} = {\text{90}} \times {\text{1}}{{\text{0}}^{ - 3}}{\text{ mol}} \times {\text{40}}{\text{.0 g/mol = 3}}{\text{.2 g}}\]
\[{\text{Mass of NaOH}} = {\text{3}}{\text{.6 g}}\]
Thus, grams of \[{\text{NaOH}}\] required to prepare $200{\text{ mL}}$ of a $0.450{\text{ M}}$ solution are \[{\text{3}}{\text{.6 g}}\].

Note:Solution is a mixture of solute and solvent. In solution, the solute is the substance that is present is less amount while the solvent is the substance that is present in a large amount. Mostly water is used as a solvent to prepare the solution. Here \[{\text{NaOH}}\] is the solute. Molarity is one of the units of concentration.