A standard solution was prepared by dissolving 0.2512 (+,- 0.0004) g of sodium t

Question

A standard solution was prepared by dissolving 0.2512 (+,- 0.0004) g of sodium thiosulfate (Na2S2O3) in distilled water. The solution was transferred to a 250-ml volumetric flask and brought to volume. A 5-ml portion of this standard was then transferred to a 50-ml volumetric flask and again brought to volume with distilled water to create the final solution.

A. Determine the concentration of Na2S2O3 in molarity (mol/L) in the final solution

B. Calculate the absolute uncertainty of the concentration of Na2S2O3 in the final solution. Assume the molar mass of Na2S2O3 has an uncertainty of +,- 0.03 g/mol.

Explanation / Answer

A) Amount of sodium thiosulphate = 0.2512 +/- 0.0004

Volume = 250mL

volume of sodium thiosulphate taken = 5mL

Volume make up to 50mL

1) moles of sodium thiosulphate = Mass / Molecular weight = 0.2512 / 158.11 = 0.001588 moles

Molarity of stock solution = moles of solute / Volume in litres = 0.001589 / 0.25 = 0.006356 molar

The volume taken is 5mL(v1) of 0.006356 molar solution(M1)

diluted to 50mL(V2) so new concentration will be (M2)

M1V1 = M2V2

M2 = M1V1 / V2 = 0.006356 X 5 / 50 = 0.0006356 (final solution) Or 0.00064 (two significant figures)

B)

If molar mass has uncertainity = +/- 0.03

And the uncertainity in mass = +/- 0.0004

Assuming no uncercatinity in measurement of volume, so uncertainity will be in moles

As the moles are obtained by dividing mass with mol wt so the uncertainity will add

Total uncertainity = +/ - 0.00000283

Uncertainity = ((0.2512)^2 / (0.0004)^2 + (158.11 / 0.03)^2 )1/2

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