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X moli de KOH se dizolva în 10x moli de apa. Calculati concentratia procentuala masica si compozitia procentuala elementara a solutiei astfel obtinuta.

Răspuns :

MKOH=39+16+1=56g/mol
KOH + H2O -> KOOH + H2
nr moli=md/M =>md=n*N=x*56=56xg
nr moli=mH2O/M => mH2O=n*M=10x*18=180xg H2O
ms=md+mH2O=56x+180x=236xg
Cp=md/ms*100=56x/236x*100=23.728%
MKOOH=39+16*2+1=72g/mol
%K=mK/M*100=39/72*100=54.16
%O=mO/M*100=32/72*100=44.44
%H=mH/M*100=1.38