Question

Q.11. Calculate the mass of NaBr needed to prepare a 50.00-mL aqueous solution that would yield the same conductivity as the NaCl (aq) solution. Show your calculation to your instructor before making the solution and measuring its conductivity.

Answer 1

Answer:

The mass of NaBr needed is 0.22969 g.

Explanation:

1 mole of NaBr contains 22.4 dm^3 of NaBr

Therefore, 0.05 dm^3 (50/1000 = 0.05 dm^3) of NaBr = 0.05/22.4 = 0.00223 mol of NaBr

MW of NaBr = 23 + 80 = 103 g/mol

Mass of NaBr = number of moles × MW = 0.00223 × 103 = 0.22969 g