Question

calculeaza partea de masa a alcoolului etilic c2h5oh in soutia obtinuta la amestecarea a 1l de alcool etilic de 96% (cu densitatea de 0.79 g/,l) si 1,972 l de apa

Answer 1

ro,solutie= 0,79g/ml

ro=m/v---> m,s= 1000mlx0,79g/ml=790g

c1=md1x100/ms1---> md1== 96x790/100g==758,4g alcool

in solutia diluata cu apa,

md2=md1= 758,4g

ms2=ms1+m,apa       ro,apa= 1g/ml---> m,apa=1972g

ms2=  790+1972g=2762g

c2= 758,4x100/2762=27,5%