
How many moles are in 4 liters of 4 M NaOH?
Answer
522.6k+ views
Hint: To solve this question, we first need to understand what is a mole. Mole is the SI unit of measurement and is used to determine the amount of a substance. A mole of any substance has exactly $6.022\times {{10}^{23}}$ particles which can be ions, atoms, electrons, or molecules.
Complete step-by-step answer:
We know that the molarity of a solution is defined as the concentration of a solute in a solution is in terms of the amount of solute per unit volume, or the molar concentration. The SI unit for molarity is mol/${{m}^{3}}$ but it is usually expressed in mol/$d{{m}^{3}}$ which is equivalent to mol/L. It can also be expressed by the capital letter M.
Molarity can be given by
\[M=\dfrac{n}{V}\]
Where M is the molarity of the solution,
V is the volume of the solution,
And n is the number of moles of solute present in the solution.
So, the number of moles of solute in a solution can be given by
\[n=M\times V\]
Now, it is given to us that the molarity of the sodium hydroxide solution M=4 mol/L, and the volume of the solution is V=4L
So, the number of moles of NaOH present in the NaOH solution will be
\[\begin{align}
& n=4\times 4 \\
& n=16mol \\
\end{align}\]
Hence, there are 16 moles of NaOH in 4 L of a 4 M NaOH solution.
Additional Information:
When some amount of a solute is dissolved in some amount of a solvent, the molarity of the resultant solution can be given by
\[M=\dfrac{{{w}_{s}}}{{{m}_{s}}}\times \dfrac{1000}{V}\]
Where, ${{w}_{s}}$= mass of the solute dissolved (in grams),
${{m}_{s}}$= molar mass of the solute dissolved (in grams),
And V= total volume of the solution.
Note: It should be noted that since there are 16 moles of NaOH present in the solution, we can calculate the mass of NaOH present.
We know that the molar mass of NaOH is 39.997 g/mol. So
\[\begin{align}
& {{w}_{s}}={{m}_{s}}\times n \\
& {{w}_{s}}=39.997\times 16 \\
& {{w}_{s}}\cong 639.952g \\
\end{align}\]
Hence to make 4L of 4M NaOH solution, we need to dissolve 639.952g of NaOH in 4L solvent.
Complete step-by-step answer:
We know that the molarity of a solution is defined as the concentration of a solute in a solution is in terms of the amount of solute per unit volume, or the molar concentration. The SI unit for molarity is mol/${{m}^{3}}$ but it is usually expressed in mol/$d{{m}^{3}}$ which is equivalent to mol/L. It can also be expressed by the capital letter M.
Molarity can be given by
\[M=\dfrac{n}{V}\]
Where M is the molarity of the solution,
V is the volume of the solution,
And n is the number of moles of solute present in the solution.
So, the number of moles of solute in a solution can be given by
\[n=M\times V\]
Now, it is given to us that the molarity of the sodium hydroxide solution M=4 mol/L, and the volume of the solution is V=4L
So, the number of moles of NaOH present in the NaOH solution will be
\[\begin{align}
& n=4\times 4 \\
& n=16mol \\
\end{align}\]
Hence, there are 16 moles of NaOH in 4 L of a 4 M NaOH solution.
Additional Information:
When some amount of a solute is dissolved in some amount of a solvent, the molarity of the resultant solution can be given by
\[M=\dfrac{{{w}_{s}}}{{{m}_{s}}}\times \dfrac{1000}{V}\]
Where, ${{w}_{s}}$= mass of the solute dissolved (in grams),
${{m}_{s}}$= molar mass of the solute dissolved (in grams),
And V= total volume of the solution.
Note: It should be noted that since there are 16 moles of NaOH present in the solution, we can calculate the mass of NaOH present.
We know that the molar mass of NaOH is 39.997 g/mol. So
\[\begin{align}
& {{w}_{s}}={{m}_{s}}\times n \\
& {{w}_{s}}=39.997\times 16 \\
& {{w}_{s}}\cong 639.952g \\
\end{align}\]
Hence to make 4L of 4M NaOH solution, we need to dissolve 639.952g of NaOH in 4L solvent.
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