Question

A standardized solution that is 0.0100 M in Na+ is necessary for a Hame photometric determination of the element. How many grams of primary-standard-grade sodium carbonate are necessary to prepare 700.0 mL of this solution?

Answer 1

reaction

Na₂CO₃ + H₂O -------->2(Na+) + (CO₃^-2) + H₂O

from equation

1 mole Na₂CO₃ gives 2 moles Na⁺

moles of Na = 0.04 * 500 /1000

moles of Na = 0.02 moles

in order to make 0.02 moles

0.01 moles of Na₂CO₃ required

quantity of Na₂CO₃ required = moles * M.W

molecular weight (M.W) of Na₂CO₃ = 105.9885 g/mol

so

quantity of Na₂CO₃ required = 0.01 * 105.9885

quantity of Na₂CO₃ required = 1.059885 grams

(b)

as we know that,

in a solution molarity = number of moles of solute present in the solution / (total number of solution in liters)

therefore number of moles of solute present in the solution = molarity X total volume of solution in liter

= 0.1355 X 79.21 /1000 = 0.01073 mol

therefore number of moles of solute present in the solution = 0.01073 moles

hope this help you

give thumbs up if you like

Thank you