Question

How will you make a \[0.35M\] glucose solution? Also, prepare a $0.12M$ solution of glucose from the above stock?

Answer 1

Hint:The question is based on the concept of Molarity. And we know that molarity refers to the mass of solute present in one litre of the solution. It is basically a concentration term which is used to determine and measure the concentration of solution.

Complete step-by-step solution:In this question, we are asked to prepare $0.35M$ glucose solution. For the sake of simplicity let us consider the volume of the solution as unity.
We can prepare a solution of glucose with desired molarity, by dissolving a certain mass of glucose in the water. So, firstly we need to calculate how much mass is required which when mixed in water gives a solution with $0.35M$.
We know that molarity (M) of a solution is given by the formula as mentioned below
$Molarity\left(M\right)=\frac{{{m}_{solute}}}{Volume\,of\,solution\,\left( L \right)}$
By substituting the value of $M=0.35M\,and\,V=1L$in the above equation, we get;
$\begin{align}
  & {{m}_{solute}}=\,Molarity\left( M \right)\times Volume\left( V \right) \\
 & {{m}_{solute}}=\,0.35\times 1=\,0.35g \\
\end{align}$
Hence, we conclude that when $0.35$ glucose is dissolved in one litre of water, then we can obtain glucose solution with molarity of $0.35M$
Now, in this question we are asked to prepare $0.12M$ Solution from the above solution of glucose.
In molarity, we know a relation which is given as- ${{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}$
Where ${{M}_{1}}=$initial molarity
${{M}_{2}}=$final molarity
${{V}_{1}}=$Initial volume
${{V}_{2}}=$ volume of new solution
For this question we know that,
$\begin{align}
  & {{M}_{1}}=0.35M,\,\,{{V}_{1}}=1L \\
 & {{M}_{2}}=0.12M,\,{{V}_{2}}=\,? \\
\end{align}$
So, we need to find out ${{V}_{2}}$
$\begin{align}
  & \therefore {{V}_{2}}=\frac{{{C}_{1}}\times {{V}_{1}}}{{{C}_{2}}} \\
 & \Rightarrow {{V}_{2}}=\frac{0.35\times 1}{0.12} \\
 & \Rightarrow {{V}_{2}}=2.99L\,\simeq 3L \\
\end{align}$
So, we see that $3L$ volume of solution is needed to prepare $0.12M$ of solution. The original volume was $1L$. So, we need to add $2L$ more of water in the original solution to prepare $3L$ of final solution and in turn molarity of $0.12M$

Note:We should note that one concentration of given solution can be converted into another concentration by simply dilution or in simple words adding more amount of solvent to obtain the desired results. Always take the volume in liters in case of molarity.