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NH2COONH4 when heated to 200 degree C gives a mixture of NH3 and CO2 vapours with a density of 16.0.
What is the degree of dissociation of NH2COONH4?
ans-1
please tell how?
1
it is 1 because NH2COONH4 when heated to 200 degree C gives a mixture of NH3 and CO2 vapours with a density of 16.0
it is 1 because
NH2COONH4 when heated to 200 degree C gives a mixture of NH3 and CO2 vapours with a density of 16.0
NH2COONH4 --------> 2NH3 + C02
t=0 a 0 0
t=t a(1-k) 2ak ak
where k is the degree of diss. a is initial conc.
Now we know that vant hoff factor is given by total conc. at time ''t'' divided by total initial conc. i.e conc. of reactant initially. thus,
i = [a(1-k) + 2ak + ak]/a
i = [a - ak +2ak + ak]/a
i = a(1+2k)/a
i = 1+ 2k
Now for strong electrolytes vant hoff factor ''i'' is equals to sum of no. of ions in the product.
therefore total no. of ions are = 2 +1 = 3
3 = 1 +2k
2k = 2
k =1
Hence the degree of dissociation for NH2COONH4 is equals to ''1''.
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