Question

How can you prepare $0.250\,L$ of $0.085\,M $ potassium dichromate$?$

Answer 1

Hint: $M$ generally refers to the molarity of a solution which can be given by, $Molarity\, = \,\dfrac{{No.\,of\,moles\,of\,solute}}{{Volume\,of\,solution\,\left( {in\,L} \right)}}$. In the question molarity of the solution and the volume of solution is given. Convert the units as required and find the number of moles of the solute i.e. potassium dichromate hence find the amount of solute required to prepare the solution.

Complete step-by-step answer:The symbol $M$ mentioned in the question refers to molarity. Molarity is defined as the number of moles of solute per $L$ of solution.
Hence the molarity can be expressed as,
$Molarity\,\, = \,\,\dfrac{{No.\,of\,moles\,of\,solute}}{{Volume\,of\,solution\,\left( {in\,L} \right)}}............\left( 1 \right)$
Now, we are provided with a ${K_2}C{r_2}{O_7}$ solution where the solute is \[{K_2}C{r_2}{O_7}\].
The volume of the solution $ = \,0.250\,L$
The molarity of the \[{K_2}C{r_2}{O_7}\] solution $ = \,0.085\,M$.
Now, the equation $\left( 1 \right)$ can be written as,
$No.\,of\,moles\,of\,{K_2}C{r_2}{O_7}\, = \,\,Molarity \times Volume\,of\,solution\,\left( {in\,L} \right)$
Putting the values we get,
\[No.\,of\,moles\,of\,{K_2}C{r_2}{O_7}\, = \,0.250\,L \times 0.085\,mol\,{L^{ - 1}}\,\, = \,\,0.02125\,mol\]
Therefore the solution contains $0.00285\,moles$ of ${K_2}C{r_2}{O_7}$.
Now, $No.\,of\,moles\;of\;a\,compound\, = \,\,\dfrac{{Given\,Weight}}{{Molecular\,Weight}}$,
Which can also be written as,
$Given\,Weight\,\, = \,\,No.\,of\,moles\,of\,a\,compound \times Molecular\,Weight$.
The molecular weight of ${K_2}C{r_2}{O_7}$ is $294.185\,g\,mo{l^{ - 1}}$
Therefore, amount of ${K_2}C{r_2}{O_7}$ required to prepare $0.250\,L$ of $0.085\,M $potassium dichromate solution $ = \,0.02125\,mol \times 294.185\,g\,mo{l^{ - 1}}\, = \,\,6.25\,g$.
Hence for preparing $0.250\,L$ of $0.085\,M$potassium dichromate solution $6.25\,g$ of potassium dichromate is properly weighed, then taken in a $250\,mL$ volumetric flask and then volume make up is done by adding distilled water.

Note: Always remember in expression of molarity the volume of the solution is in$L$. Make sure to check the units properly so that you do not mix up the SI and CGS units. Do the calculation in a stepwise manner in order to avoid errors as you can keep track of where the same units are cancelling each other if done one step at a time.