Question

In: Chemistry

Question 1: A solution of F– is prepared by dissolving 0.0712 ± 0.0005 g of NaF...

Question 1: A solution of F– is prepared by dissolving 0.0712 ± 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.

Question 20. A solution of HNO3 is standardized by reaction with pure sodium carbonate.

A volume of 27.64 ± 0.05 mL of HNO3 solution was required for complete reaction with 0.9875 ± 0.0009 g of Na2CO3, (FM 105.988 ± 0.001). Find the molarity of the HNO3 and its absolute uncertainty.

Solutions

Expert Solution

1)

The concentration of NaF = Mass of NaF X 1000 / Molecular weight of NaF X volume of water

concentration of NaF = 0.0712 X 1000 / 41.989 X 164 = 0.013 M

Now uncertainities in multiplication and division will propagate as

relative uncertainity in concentration

= [(relative uncertainity in mass)2 + (relative uncertainity in molecular weight)2 + (relative uncertaintiy in volume)2] 1/2

Relative uncertainity in concentration = [(0.0005 / 0.0712)2 + (0.001/41.989)2 + (0.07/164)2]1/2

Relative uncertainity = [ (0.000049 + 5.66X10^-10 + 18.14X10^-8 ) 1/2 = 0.007

Absolute uncertainity in concentration = Relative uncertainity X actual value = 0.007 X 0.013 = 0.000091

2) the reaction of acid with the given base will be

2HNO3 + Na2CO3 --> 2NaNO3 + H2O + CO2

As per equation the moles of HNO3 will be twice that used of Na2CO3

(Molarity X volume ) HNO3 = 2X (moles) Na2CO3

Moles of Na2CO3 = Mass / Molecular weight = 0.9875 / 105.988 = 0.00932

relative uncertainity in moles of Na2CO3 = [(0.0009/0.9875)2 + (0.001/105.988)2]1/2

Relative uncertainity in moles of Na2CO3 = (83.046 X 10^-8 + 89.019 X 10^-12]1/2

Relative uncertainity in moles of Na2CO3 = 0.000911

(Molarity X volume ) HNO3 = 2X (moles) Na2CO3 = 2 X 0.00932

molarity = 0.674 Molar

Relative uncertainity in molarity = [relative uncertaintiy in moles )2 + (relative uncertainity in volume)2]1/2

Relative uncertainity in molarity = [(0.000911)2 + (0.05/27.64)2 )1/2

Relative uncertainity in molarity =    [8.2X10^-7 +3.24 X 10^-6]1/2 = 0.002

So absolute uncertainity in molarity = Relative uncertainity in molarity X molarity = 0.002 X 0.674 = 0.00135


Related Solutions

A solution is made by a dissolving 16.8 g of sodium fluoride , NaF , in...
A solution is made by a dissolving 16.8 g of sodium fluoride , NaF , in enough water to make exactly 500.ml of solution . Calculate the molarity of each species: NaF, Na+ and F-.
a) What is the molarity of the solution that was prepared by dissolving 3.25 g of...
a) What is the molarity of the solution that was prepared by dissolving 3.25 g of sulfuric acid in water to a total volume of 500.0 mL? b)What is the molarity of the hydrogen ion in part a if you assume the sulfuric acid ionizes completely? Write a balanced chemical equation.
A solution is prepared by dissolving 23.7 g of CaCl2 in 375 g of water. The...
A solution is prepared by dissolving 23.7 g of CaCl2 in 375 g of water. The density of the resulting solution is The mole fraction of Cl- in this solution is __________ M.
A solution is prepared by dissolving 23.7 g of CaCl2 in 375 g of water. The...
A solution is prepared by dissolving 23.7 g of CaCl2 in 375 g of water. The density of the resulting solution is 1.05 g/mL. The concentration of CaCl2 in this solution is M
A solution is prepared by dissolving 32.5 g of a nonvolatile solute in 200 g of...
A solution is prepared by dissolving 32.5 g of a nonvolatile solute in 200 g of water. The vapor pressure above the solution is 21.85 Torr and the vapor pressure of pure water is 23.76 Torr at this temperature. What is the molecular weight of the solute?
A solution was prepared by dissolving 29.0g KCl in 225 g of water. 1) Calculate the...
A solution was prepared by dissolving 29.0g KCl in 225 g of water. 1) Calculate the mass percent of KCl in the solution. 2)Calculate the mole fraction of the ionic species KCl in the solution. 3) Calculate the molarity of KCl in the solution if the total volume of the solution is 239 mL. 4) Calculate the molarity of KCl in the solution.
A solution of an unknown acid is prepared by dissolving 0.250 g of the unknown in...
A solution of an unknown acid is prepared by dissolving 0.250 g of the unknown in water to produce a total volume of 100.0 mL. Half of this solution is titrated to a phenolphthalein endpoint, requiring 12.2 mL of 0.0988 M KOH solution. The titrated solution is re-combined with the other half of the un-titrated acid and the pH of the resulting solution is measured to be 4.02. What is are the Ka value for the acid and the molar...
A solution is prepared by dissolving 29.2 g of glucose (C6H12O6) in 355 g of water....
A solution is prepared by dissolving 29.2 g of glucose (C6H12O6) in 355 g of water. The final volume of the solution is 378 mL . For this solution, calculate each of the following. molarity molality percent by mass mole fraction mole percent
A solution was prepared by dissolving 26.0 g of KCl in 225 g of water. Part...
A solution was prepared by dissolving 26.0 g of KCl in 225 g of water. Part A: Calculate the mole fraction of KCl in the solution. Part B: Calculate the molarity of KCl in the solution if the total volume of the solution is 239 mL. Part C: Calculate the molality of KCl in the solution.
A solution of sucrose is prepared by dissolving 0.5 g in 100 g of water. Calculate:...
A solution of sucrose is prepared by dissolving 0.5 g in 100 g of water. Calculate: a. Percent weight in weight b. The molal concentration of sucrose and water c. The mole fraction of sucrose and water in the solution
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT