A sample of Co(NH) Cl, was analyzed by \"precipitation titration\" with silver n
ID: 542355 • Letter: A
Question
Explanation / Answer
from equation,
1 mol Co(NH3)6Cl3 = 3 mol AgNO3
mass of Agno3 IN 0.5 L AgNO3 = 0.8114 g
molarity of Agno3 solution = (w/mwt)*(1/v in L)
= (0.8114/169.87)*(1/0.5) = 0.00955 M
volume(ml) of AgNo3 used = 13.08-0.89 = 12.19 ml
moles of AgNo3 used in titration = V*M/1000
= 12.19*0.00955/1000
= 0.0001164 mol
moles of Co(NH3)6Cl3 present in sample = 0.0001164*1/3 = 3.88*10^-5 mol
mass of Co(NH3)6Cl3 present in sample = n*Mwt
= 3.88*10^-5*267.48
= 0.0104 g
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