
The amount of \[{\text{NaOH}}\] in \[{\text{750 ml}}\] of \[0.2{\text{ M}}\] solution (Molecular weight \[ = 40\] ) is:
A. \[{\text{2 gm}}\]
B. \[{\text{4 gm}}\]
C. \[6{\text{ gm}}\]
D. \[8{\text{ gm}}\]
Answer
572.1k+ views
Hint:The molarity of a solution gives the number of moles of a solute present in one liter of solution.
\[{\text{Molarity }} = {\text{ }}\dfrac{{{\text{Mass of solute}}}}{{{\text{molar mass of solute}}}} \times {\text{ }}\dfrac{{1000}}{{{\text{volume of solution in mL}}}}\]
When you multiply molarity with volume in liter, you get the number of moles of solute. When you multiply the number of moles of solute with molar mass, you get the mass of solute.
Complete answer:
\[{\text{1 M}}\] solution contains \[40{\text{ grams}}\] of \[{\text{NaOH}}\] present in \[{\text{1 L}}\] or \[{\text{1,000 mL}}\] of solution.
Write the formula to calculate the molarity of the solution
\[{\text{Molarity }} = {\text{ }}\dfrac{{{\text{Mass of solute}}}}{{{\text{molar mass of solute}}}} \times {\text{ }}\dfrac{{1000}}{{{\text{volume of solution in mL}}}}\]
Rearrange above formula to calculate the mass of solute
\[{\text{Mass of solute }} = {\text{ }}\dfrac{{{\text{Molarity }} \times {\text{ molar mass of solute }} \times {\text{ volume of solution in mL}}}}{{{\text{1000}}}}\]
In the above equation, the solute is sodium hydroxide.
Rewrite above equation in terms of sodium hydroxide
\[{\text{Mass of NaOH }} = {\text{ }}\dfrac{{{\text{Molarity }} \times {\text{ molar mass of NaOH }} \times {\text{ volume of solution in mL}}}}{{{\text{1000}}}}\]
Substitute values in the above expression to calculate the mass of sodium hydroxide
\[\Rightarrow {\text{Mass of NaOH }} = {\text{ }}\dfrac{{{\text{0}}{\text{.2 M }} \times {\text{ 40 g/mol}} \times {\text{ 750 mL}}}}{{{\text{1000}}}}\]
\[\Rightarrow {\text{Mass of NaOH}} = 6{\text{ g}}\]
Hence, the mass of sodium hydroxide present in \[{\text{750 ml}}\] of \[0.2{\text{ M}}\] solution is \[6{\text{ g}}\] .
Hence, the correct option is the option (C).
Additional information:
You can express the concentration of solute in a solution in several different ways. Some of these include molarity, molality, percent by mass, percent by volume, weight by volume percent, mole fraction, normality and parts per million. You can use a certain formula to convert one form of concentration term into another form of concentration term.
Note:
Similar to molarity, another related concept is that of molality. Molality gives the number of moles of solute present in one kilogram of solvent. Thus, \[{\text{1 m}}\] solution contains \[40{\text{ grams}}\] of \[{\text{NaOH}}\] present in \[{\text{1 kg}}\] or \[{\text{1,000 g}}\] of water.
\[{\text{Molarity }} = {\text{ }}\dfrac{{{\text{Mass of solute}}}}{{{\text{molar mass of solute}}}} \times {\text{ }}\dfrac{{1000}}{{{\text{volume of solution in mL}}}}\]
When you multiply molarity with volume in liter, you get the number of moles of solute. When you multiply the number of moles of solute with molar mass, you get the mass of solute.
Complete answer:
\[{\text{1 M}}\] solution contains \[40{\text{ grams}}\] of \[{\text{NaOH}}\] present in \[{\text{1 L}}\] or \[{\text{1,000 mL}}\] of solution.
Write the formula to calculate the molarity of the solution
\[{\text{Molarity }} = {\text{ }}\dfrac{{{\text{Mass of solute}}}}{{{\text{molar mass of solute}}}} \times {\text{ }}\dfrac{{1000}}{{{\text{volume of solution in mL}}}}\]
Rearrange above formula to calculate the mass of solute
\[{\text{Mass of solute }} = {\text{ }}\dfrac{{{\text{Molarity }} \times {\text{ molar mass of solute }} \times {\text{ volume of solution in mL}}}}{{{\text{1000}}}}\]
In the above equation, the solute is sodium hydroxide.
Rewrite above equation in terms of sodium hydroxide
\[{\text{Mass of NaOH }} = {\text{ }}\dfrac{{{\text{Molarity }} \times {\text{ molar mass of NaOH }} \times {\text{ volume of solution in mL}}}}{{{\text{1000}}}}\]
Substitute values in the above expression to calculate the mass of sodium hydroxide
\[\Rightarrow {\text{Mass of NaOH }} = {\text{ }}\dfrac{{{\text{0}}{\text{.2 M }} \times {\text{ 40 g/mol}} \times {\text{ 750 mL}}}}{{{\text{1000}}}}\]
\[\Rightarrow {\text{Mass of NaOH}} = 6{\text{ g}}\]
Hence, the mass of sodium hydroxide present in \[{\text{750 ml}}\] of \[0.2{\text{ M}}\] solution is \[6{\text{ g}}\] .
Hence, the correct option is the option (C).
Additional information:
You can express the concentration of solute in a solution in several different ways. Some of these include molarity, molality, percent by mass, percent by volume, weight by volume percent, mole fraction, normality and parts per million. You can use a certain formula to convert one form of concentration term into another form of concentration term.
Note:
Similar to molarity, another related concept is that of molality. Molality gives the number of moles of solute present in one kilogram of solvent. Thus, \[{\text{1 m}}\] solution contains \[40{\text{ grams}}\] of \[{\text{NaOH}}\] present in \[{\text{1 kg}}\] or \[{\text{1,000 g}}\] of water.
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