Capacity in mLNBS Tolerances for Class A Volumetric Ware (mL) volumetric flasks

Question

Capacity in mLNBS Tolerances for Class A Volumetric Ware (mL) volumetric flasks ansfer pipets burets 5 10 25 50 100 200 250 0.02 0.03 0.04 0.08 0.10 0.12 0.006 0.01 0.02 0.03 0.05 0.02 0.03 0.05 Clearly, the larger the volume and the amount of material used, the smaller the uncertainty in knowing the concentration of the final solution One further issue remains to be considered in the experimental design. At the same time as scientists consider the accuracy and precision of their measurements, proper practice mandates that waste be minimized. This is essential for hazardous materials. Thus, you must always determine in advance how to make the minimum amount of the solution of the required concentration and precision Using an analytical balance that has a sensitivity of ±0.3 mg and any of the volumetric flasks in the table above, design a procedure to prepare at least 50 mL of a 3.4x10 2 M primary standard solution of lithium fluoride with a precision of 0.5%. Assume the lithium fluoride is pure and dry

Explanation / Answer

Steps to prepare 3.4 X 102 M primary standard solution of LiF

1. As we have to prepare 50 ml of this solution so, we have to take clean and dry volumetric flask of 50 mL

2. To prepare 3.4 X 102 M solution, we have use formula given below to calculate the amount of amount of LiF taken

as, molarity (M)=number of moles (n)/volume of solution (V) in litres

M=n/V

M=mass of LiF taken (m) /mol. weight of LiF (M) X vol of solution prepared

so, 3.4 X 102 M=m/25.9392 g/molX 0.05 L

m=340 M X 25.9392 g/mol X 0.05 L

m=440.9664 g

to make the solution with required accurary and precision, we have to prepare higher volume (around 500- 1000 ml of the solution using required weight of LiF as per formula and take out the 50 ml of solution.

other way is to prepare solution of higher concentrartion or molarity and then dilute it to required molarity i.e 3.4X 102 M

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