To prepare a very dilute solution, it is advisable to perform successive dilutio

Question

To prepare a very dilute solution, it is advisable to perform successive dilutions of a single prepared reagent solution, rather than to weight out a very small mass or to measure a very small volume of stock chemical. A solution was prepared by transferring 0.661g of K2Cr2O7 to a 250.0-mL volumetric flask and adding water to the mark. A sample of this solution of volume 1.00-mL was transferred to a 500.0-mL volumetric flask and diluted to the mark with water. Then 10.0-mL of the diluted solution was transferred to a 250.0-mL flask and diluted to the mark with water. (a) What is the final concentration of K2Cr2O7 in solution? What mass of K2Cr2O7 is in the final solution? (The answer to the last question gives the amount that would have had to have been weighed out if the solution had been prepared directly.)

Explanation / Answer

molar mass of K2Cr2O7=294

moles of K2Cr2O7=0.661/294=0.002248

molarity in first solution= (0.002248) / (250/1000)=0.00899mol/l

so 1 ml will have moles of K2Cr2O7= [8.99* 10^-3 /1000] =8.99* 10^-6moles

now this is transferred to 500 ml solution so molarity of new solution= no of new moles/ volume=

=(8.99* 10^-6)/0.5l =1.798*10-5

now 10 ml of this solution will have  moles of K2Cr2O7= [1.798*10-5*10 /1000] =1.798*10-7moles

now this is transferred to 250 ml solution so molarity of new solution= no of new moles/ volume=

=(1.798*10-7)/0.25l =7.192*10-7 mol/l

mass of K2Cr2O7 in final solution= 1.798*10-7* 294= 5.286*10-5 gm left

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