A solution was prepared by dissolving 116.0 g of NaCl in 3500. ml of wter. After

Question

A solution was prepared by dissolving 116.0 g of NaCl in 3500. ml of wter. After this solution was prepared, a 150.00 ml was pipetted out and transfered to a 200 ml volumetric flask and diluted to the mark with wter. From this last solution a 75.0 ml sample was transfered into a 100.0 ml volumetric flask and diluted to the mark with water. Calculate the concentration of the last solution. How could this solution be prepared directly from the first (stock solution) ?

Clear explaination is greatly apprecited. Thanks!

Explanation / Answer

Concentration in initial solution

Molarity = weight of solute in g x 1000 / (molecular weight of solute x volume of solution in mL)

Here solute is NaCl (mol. weight = 58.5) and solvent is water

Molarity = 116.0 x 1000 / 58.5 x 3500 = 0.57 M

After a 150.00 ml was pipetted out and transfered to a 200 ml volumetric flask and diluted to the mark with water, the concentration will be

V1S1=V2S2

=>S2 = 150 x 0.57 / 200 =0.43 M

Again,  a 75.0 ml sample was transfered into a 100.0 ml volumetric flask and diluted to the mark with water, the concentration will be

V1S1=V2S2

=> S2 = 75 x 0.43 / 100 = 0.32 M

Hence, the concentration of the last solution is 0.32 M

For making 0.32 M solution directly from stock solution in 100 mL volumetric flask

V1S1=V2S2

=>V1= 100 x 0.32 / 0.57 = 56.14 mL

Take 56.14 mL of stock solution in a 100 mL of volumetric flask and make up the volume upto the mark, which will give you 100 mL of 0.32 M solution

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