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Question: How many grams of \[{\text{NaOH}}\] are required to prepare \(200{\text{ mL}}\) of a \(0.450{\text{ ...

How many grams of NaOH{\text{NaOH}} are required to prepare 200 mL200{\text{ mL}} of a 0.450 M0.450{\text{ M}} solution?

Explanation

Solution

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

Complete solution:
We have to calculate the mass of NaOH{\text{NaOH}} in 200 mL200{\text{ mL}} of a 0.450 M0.450{\text{ M}} solution.
Using the given molar concentration and volume of the solution calculate the number of moles of NaOH{\text{NaOH}} as follows:
The equation of molarity is as follows:
Molarity(M)=Number of moles of solute(mol)Volume of solution(L){\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:
Number of moles of solute(mol)=Molarity(M)×Volume of solution(L){\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 mL=200×10−3 L200{\text{ mL}} = 200 \times {10^{ - 3}}{\text{ L}} for the volume of the solution, 0.450 M0.450{\text{ M}} for the molarity. Thus,
Number of moles of NaOH=0.450 M×200×10−3 L{\text{Number of moles of NaOH}} = {\text{0}}{\text{.450 M}} \times 200 \times {10^{ - 3}}{\text{ L}}
Number of moles of NaOH=90×10−3 mol{\text{Number of moles of NaOH}} = 90 \times {10^{ - 3}}{\text{ mol}}
Thus, the number of moles of NaOH{\text{NaOH}} in 200 mL200{\text{ mL}} of a 0.450 M0.450{\text{ M}} solution are 90×10−3 mol90 \times {10^{ - 3}}{\text{ mol}}.
Now, we can convert these calculated moles of NaOH{\text{NaOH}} to mass as follows:
The mathematical equation related to moles, mass and molar mass is as follows:
Number of moles of solute(mol)=Mass(g)Molar mass(g/mol){\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:
Mass(g)=Number of moles of solute(mol)×Molar mass(g/mol){\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 90×10−3 mol90 \times {10^{ - 3}}{\text{ mol}} for moles of NaOH{\text{NaOH}} , 40.0 g/mol40.0{\text{ g/mol}} for molar mass of NaOH{\text{NaOH}}. Thus,
Mass of NaOH=90×10−3 mol×40.0 g/mol = 3.2 g{\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}}
Mass of NaOH=3.6 g{\text{Mass of NaOH}} = {\text{3}}{\text{.6 g}}
Thus, grams of NaOH{\text{NaOH}} required to prepare 200 mL200{\text{ mL}} of a 0.450 M0.450{\text{ M}} solution are 3.6 g{\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 NaOH{\text{NaOH}} is the solute. Molarity is one of the units of concentration.