Question

How many moles of solute is contained in 1.75 L of 0.215 M solution?

Answer 1

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 solution M=0.215 mol/L, and the volume of the solution is V=1.75L
So, the number of moles of solute present in the sample solution will be
\[\begin{align}
  & n=0.215\times 1.75 \\
 & n\cong 0.376mol \\
\end{align}\]
Hence, there are approximately 0.376 moles of solute in 1.75 L of a 0.215 M solution.

Note: 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.