A solution of F– is prepared by dissolving 0.0710 ± 0.0005 g of NaF (molecular w

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A solution of F– is prepared by dissolving 0.0710 ± 0.0005 g of NaF (molecular weight = 41.989 ± 0.001 g/mol) in 164.00 ± 0.07 mL of water. Calculate the concentration of F– in solution and its absolute uncertainty.

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Explanation / Answer

weight of NaF = 0.0710 + 0.0005 = 0.0715gm

volume of water                 = 164.00+0.07 = 164.07ml
molecular weight                = 41.989 +0.001 =41.99g/mole

molarity = weight of NaF*1000/Gram molar mass of NaF * volume in ml
         = 0.0715*1000/41.99*164.07
         = 71.5/6889.2993
         = 0.01037 M

NaF -----> Na+ + F^-
0.01037M         0.01037M
concentration of F- is = 0.01037±0.00003M

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