[H|D]2O + Cl2 = [H|D]Cl
[H|D]Cl + AgNO3 = AgCl + [H|D]NO3
Molar mass H2O: W1 = 18.01532 g/mol
Molar mass D2O: W2 = 20.02763 g/mol
Molar mass AgCl: W3 = 143.3214 g/mol
Sample mass: Ms = 1g
Precipitate: Mp = 0.3800g
Assay Volume: Vs = 1.00L
Aliquot Volume: Va = 25.00mL
Let n1 and n2 be the amounts of hydrogen and deuterium in the 1g mixed-hydrogen sample.
Ms = n1 W1 + n2 W2
or
1g = 18.01532 n1 + 20.02763 n2
The mole amount of silver chloride precipitate equals the mole amount of mixed-hydrogen in the aliquot, which is a representative fraction of that in the original sample.
(W3/Mp) = (n1+n2)(Va/Vs).
or
(143.3214/0.3800) = (n1+n2)(0.02500/1.00)
Solve these simultaneous equations for n1 and n2.
Mass % of D2O is then: (100% n2 W2 / Ms)
A1g sample of enriched water, a mixture of H2O AND D2O, reacted completely with Cl2 to give a mixture of HCl and DCl. The HCl and DCl were then dissolved in pure H2O to make a 1.00 L solution. A 25.00 mL, sample of the 1 L solution was reacted with excess AgNO3 and 0.3800g of an AgCl precipitate formed. What was the mass % of D2O in the original sample of enriched water?
2 answers
46.0%
1 answer