Question

A solution of F− is prepared by dissolving 0.0722±0.0005 g NaF (molar mass = 41.989±0.001 g/mol) in 154.00±0.06 mL of water. Calculate the concentration of F− in solution and its absolute uncertainty.

Significant figures are graded for this problem. To avoid rounding errors, do not round your answers until the very end of your calculations.

Answer 1

Mass of F^- added (as NaF) = 0.07220.0005 g.

Molar mass of NaF = 41.9890.001 g.

Mole(s) of NaF corresponding to 0.0722 g = (mass of NaF)/(molar mass of NaF)

= (0.0722 g)/(41.989 g)

= 0.0017195 mole.

Volume of water = 154.000.06 mL.

Molarity of F^- = (moles of F^-)/(volume of water in L)

= (0.0017195 mole)/[(154.00 mL)*(1 L)/(1000 mL)]

= (0.0017195 mole)/(0.154 L)

= 0.01116558 mol/L

Relative uncertainty in the mass of F^-, U₁ = (0.0005 g)/(0.0722 g)

= 0.0069252

Relative uncertainty in the molar mass of NaF, U₂ = (0.001 g/mol)/(41.989 g/mol)

= 2.381576*10^-5

Relative uncertainty in the volume of water, U₃ = (0.06 mL)/(154.00 mL) = 3.89610*10^-4.

Relative uncertainty in the molarity of F^-

= √U₁² + U₂² + U₃²

= √(0.0069252)² + (2.381576*10^-5)² + (3.89610*10^-4)²

= 0.00693619

Absolute uncertainty in the molarity of F^- = (relative uncertainty)*(molarity of F^-)

= (0.00693619)*(0.01116558 mol/L)

= 7.744658*10^-5 mol/L = 0.00007744658 mol/L

All the readings have 5 sig. digits while the uncertainties have only one sig. fig. Therefore, the molarity of F^- = 0.0111650.00008 mol/L (ans).