A solution of F– is prepared by dissolving 0.0946 ± 0.0005 g of NaF (molecular w

Question

A solution of F– is prepared by dissolving 0.0946 ± 0.0005 g of NaF (molecular weight = 41.989 ± 0.001 g/mol) in 154.00 ± 0.07 mL of water. Calculate the concentration of F– in solution and its absolute uncertainty.

A solution of F is prepared by dissolving 0.0946 t 0.0005 g of NaF (molecular weight 41.989 t 0.001 g/mol) in 154.00 t 0.07 mL of water. Calculate the concentration of F in solution and its absolute uncertainty Note: Significant figures are graded Number Number for this problem. To avoid rounding errors, do not round your answers until the very end of your calculations.

Explanation / Answer

[F-] = mol/V

mol = mass/MW

so

[F-] = mass/(MW*V)

for divisions, we must get relative uncertainties, then add them later

[F-] = (0.0946)/(41.989*154/1000) = 0.014629 (use 3 sig fig, since the leas tsig fig number is 3)

[F-] = 0.0146 M

for uncertainty:

mass = 0.0005/0.0946 * 100 = 0.52854

MW = 0.001/41.989 * 100 = 0.002381

V = 0.07/154 * 100 = 0.045454

Total = 0.52854+0.002381+0.045454 = 0.576375

which...

%0.576375 final error:

0.576375/100*0.0146 = 0.000084150 M uncertainty

so

[F-] = 0.0146 M +/- 0.0000842

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