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Question: How can you prepare \(0.250\,L\) of \(0.085\,M \) potassium dichromate\(?\)...

How can you prepare 0.250 L0.250\,L of 0.085 M0.085\,M potassium dichromate??

Explanation

Solution

MM generally refers to the molarity of a solution which can be given by, Molarity = No. of moles of soluteVolume of solution (in L)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 MM mentioned in the question refers to molarity. Molarity is defined as the number of moles of solute per LL of solution.
Hence the molarity can be expressed as,
Molarity  =  No. of moles of soluteVolume of solution (in L)............(1)Molarity\,\, = \,\,\dfrac{{No.\,of\,moles\,of\,solute}}{{Volume\,of\,solution\,\left( {in\,L} \right)}}............\left( 1 \right)
Now, we are provided with a K2Cr2O7{K_2}C{r_2}{O_7} solution where the solute is K2Cr2O7{K_2}C{r_2}{O_7}.
The volume of the solution = 0.250 L = \,0.250\,L
The molarity of the K2Cr2O7{K_2}C{r_2}{O_7} solution = 0.085 M = \,0.085\,M.
Now, the equation (1)\left( 1 \right) can be written as,
No. of moles of K2Cr2O7 =  Molarity×Volume of solution (in L)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 K2Cr2O7 = 0.250 L×0.085 mol L−1  =  0.02125 molNo.\,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 moles0.00285\,moles of K2Cr2O7{K_2}C{r_2}{O_7}.
Now, No. of moles  of  a compound =  Given WeightMolecular WeightNo.\,of\,moles\;of\;a\,compound\, = \,\,\dfrac{{Given\,Weight}}{{Molecular\,Weight}},
Which can also be written as,
Given Weight  =  No. of moles of a compound×Molecular WeightGiven\,Weight\,\, = \,\,No.\,of\,moles\,of\,a\,compound \times Molecular\,Weight.
The molecular weight of K2Cr2O7{K_2}C{r_2}{O_7} is 294.185 g mol−1294.185\,g\,mo{l^{ - 1}}
Therefore, amount of K2Cr2O7{K_2}C{r_2}{O_7} required to prepare 0.250 L0.250\,L of 0.085 M0.085\,M potassium dichromate solution = 0.02125 mol×294.185 g mol−1 =  6.25 g = \,0.02125\,mol \times 294.185\,g\,mo{l^{ - 1}}\, = \,\,6.25\,g.
Hence for preparing 0.250 L0.250\,L of 0.085 M0.085\,Mpotassium dichromate solution 6.25 g6.25\,g of potassium dichromate is properly weighed, then taken in a 250 mL250\,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 inLL. 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.