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Question: How many grams of \({\text{NaOH}}\) are present in \[250{\text{mL}}\] of \[0.5{\text{M}}\] \({\text{...

How many grams of NaOH{\text{NaOH}} are present in 250mL250{\text{mL}} of 0.5M0.5{\text{M}} NaOH{\text{NaOH}} solution?
A. 7.32g7.32{\text{g}}
B. 3.8g3.8{\text{g}}
C. 5g5{\text{g}}
D. 0.5g0.5{\text{g}}

Explanation

Solution

Hint: Mole concept is based on how we measure entities. Molarity is considered as the concentration of a solute. It is the relative amount of solute in a solution.

Given data:
Molarity of NaOH{\text{NaOH}}, M=0.5M{\text{M}} = 0.5{\text{M}}
Volume of solutionV=250mL{\text{V}} = 250{\text{mL}}

Complete step by step solution:
One mole is the amount of a substance that contains as many particles as there are in 12g12{\text{g}} of C12{\text{C}} - 12 atoms which is exactly equal to 6.022×10236.022 \times {10^{23}} particles. This absolute number is called Avogadro number. Avogadro number tells about the number of constituent particles present in one mole of substance. Relative atomic mass of an atom is the ratio of the average mass of an atom to 112\dfrac{1}{{12}} the mass of a carbon atom. So one mole of any substance contains Avogadro's number of atoms in it. Molecular mass is the sum of atomic masses of the elements present in a molecule. Molarity refers to the amount of solutes per litre of solution. Amount of solute and volume of solvent can be calculated from molarity. Molarity is represented by M{\text{M}}. The equation for finding molarity is given below:
M=nV{\text{M}} = \dfrac{{\text{n}}}{{\text{V}}}, where n{\text{n}} is number of moles of solute
V{\text{V}} is the volume of solution in litres
Here, molarity is given.
M=0.5M,V=250mL, n = ?{\text{M}} = 0.5{\text{M}},{\text{V}} = 250{\text{mL, n = ?}}
Here we have to calculate the number of moles of NaOH{\text{NaOH}} from the above equation.
0.5M=n250mL0.5{\text{M}} = \dfrac{{\text{n}}}{{250{\text{mL}}}}
Since molarity is number of moles of solute per volume in litres, the equation can be written as
i.e., 0.5moles1000mL=n250mL\dfrac{{0.5{\text{moles}}}}{{1000{\text{mL}}}} = \dfrac{{\text{n}}}{{250{\text{mL}}}}
n=0.5moles×250mL1000mL{\text{n}} = \dfrac{{0.5{\text{moles}} \times 250{\text{mL}}}}{{1000{\text{mL}}}}
n=0.5moles4=0.125moles{\text{n}} = \dfrac{{0.5{\text{moles}}}}{4} = 0.125{\text{moles}}
Number of moles is defined as mass divided by molecular mass. Here mass (m) is referred to the weight of NaOH{\text{NaOH}}that has to be taken. Molecular mass of NaOH{\text{NaOH}} is 40g40{\text{g}}
n=m40g\therefore {\text{n}} = \dfrac{{\text{m}}}{{40{\text{g}}}}
0.125moles=m40g m=0.125moles×40g m = 5g  0.125{\text{moles}} = \dfrac{{\text{m}}}{{40{\text{g}}}} \\\ {\text{m}} = 0.125{\text{moles}} \times 40{\text{g}} \\\ {\text{m = 5g}} \\\
Therefore 5g5{\text{g}} of NaOH{\text{NaOH}} is present in 250mL250{\text{mL}} of 0.5M0.5{\text{M}} NaOH{\text{NaOH}} solution.
Hence option C is correct.

Additional information:
Molar mass is the mass of one mole of substance in grams which is equal to atomic mass. Concentration is a measure in the form of molarity. The process by which a solution is made less concentrated via addition of more solvent is called dilution.

Note: Normality is another measure which can be calculated from molarity. Molarity multiplied with acidity or basicity is called normality. Acidity is the number of hydroxyl groups which can be donated. Basicity is the number of protons which can be donated.