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Answer to Question #29530 in Inorganic Chemistry for aayushi
Question #29530
a strong current of trivalent gaseous boron passed through germanium crystal decreases the density of the crystal due to part replacement of germanium by boron and due to interstitial vacancies is created by missing germanium atoms. in one such experiment one gram germanium is taken an d boron atoms are found to be 150 ppm by weight when the density of germanium crystal decreases by 4%. calculate the percentage of missing vacancies due to germanium which are filled up by boron atoms.
Expert's answer
1 g Germanium------- 150 ppm = 1.5 * 10^-4 g Boron
4% decrease---- now weight is 0.96 g(Ge + B)----0.04 g missing
0.04 missing-------- (0.04-0.00015)g Ge missing------(0.04-0.00015)/72.61 moles
Ge missing=5.488*10^-4 moles
Boron replaced= 1.5*10^-4 g = (1.5*10^-4)/10 moles
equivalently replaced = 3/4((1.5*10^-4)/10) moles of Ge ( Valency of B is 3 and
Ge is 4)
= 0.1125*10^-4 moles of Ge
% of Missing vacancies filled by B atoms = 0.1125*100/5.488 = 2.05 %
4% decrease---- now weight is 0.96 g(Ge + B)----0.04 g missing
0.04 missing-------- (0.04-0.00015)g Ge missing------(0.04-0.00015)/72.61 moles
Ge missing=5.488*10^-4 moles
Boron replaced= 1.5*10^-4 g = (1.5*10^-4)/10 moles
equivalently replaced = 3/4((1.5*10^-4)/10) moles of Ge ( Valency of B is 3 and
Ge is 4)
= 0.1125*10^-4 moles of Ge
% of Missing vacancies filled by B atoms = 0.1125*100/5.488 = 2.05 %
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nicely explained
Dear kuldeep, The decrease of the crystal weight (4%) consists of two components: 1) loss of Ge and 2) acquiring of B which is lighter than Ge (so it is the same effect as loss). So we have two negative effects which together make a contribution of -4%. If we deduct the part of B we will get the part of Ge that was lost.
in 0.96g quantity both Ge and B are present but in 0.04 missing quantity of crystal how u calculate the weight of Ge by 0.04-0.00015
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