Question

How would you prepare $300mL$ of $0.750M$ $KCl$?

Answer 1

Hint:To prepare any solution we should have concentration of solute. The concentration of any solution can be find out by the Molarity and Molality. In our problem, we will use concept of Molarity only because in molality, the weight of solvent (in$Kg$) is used and In Molarity, the volume of the solution is used. With the help of Molarity, we can find out how much specific element or compound is used or has been dissolved in a certain amount of the solution.

Complete step-by-step answer:Molarity can be defined as the number of moles of the solute per litre or per $d{m^3}$ of the solution. The S.I. unit of Molarity is mol/litre or mol $d{m^{ - 3}}$. For e.g. $0.25M$ solution of $NaOH$ means that $0.25mol$ of $NaOH$ has been dissolved in one litre. Mathematically, it is given as: Molarity (M) $ = $ moles of solute /volume of solution
If ${w_B}g$ of the solute of molecular mass ${m_B}$ are dissolved in $V$ litre of solution.
$Molarity(M) = \dfrac{{{n_B}}}{V}$
Where,${n_B} = \dfrac{{{w_B}}}{{{m_B}}}$
The expression will become
$
  M = \dfrac{{{w_B}}}{{{m_B} \times V}} \\
  Molarity \times {m_B} = \dfrac{{{w_B}}}{V} \\
 $ In this expression, $\dfrac{{{w_B}}}{V}$ is the strength of the solution $(g/L)$.
Now, first we find out molecular mass ${m_B}$ of $KCl$.
Molecule mass of K is \[39.0983u\] and molecular mass of Cl is \[35.453u.\]
Therefore, $
  {m_B} = 1 \times 39.0983 + 1 \times 35.453 \\
   = 74.5513u \\
 $
Putting all the values in above expression. We get,
$
  M = \dfrac{{{w_B}}}{{{m_B}}} \times \dfrac{{1000}}{{V(mL)}} \\
  0.75M = \dfrac{{{w_B}}}{{74.5}} \times \dfrac{{1000}}{{300}} \\
   \Rightarrow {w_B} = \dfrac{{0.75 \times 74.5 \times 300}}{{1000}} \\
   = \dfrac{{16762.5}}{{1000}} \\
   = 16.76g \\
 $
In the laboratory, if you want to prepare this solution, first measure out $16.7g$ of $KCl$ and put it in the volumetric flask. After this add $300mL$ solvent means water to it. Solution of $KCl$ is formed.

Note:Remember that Molarity and Molality both are different. Molality is number of the solute dissolved in $1Kg$ of solvent and Molarity is number of moles of the solute dissolved per litre. To get accurate volume of the solution follows these steps in order. Keep in mind that Molarity is dependent on volume and temperature. It decreases with increase of temperature as volume increase with temperature.