ms1 = 200 g , c%1 = 49% , c%2 = 26,203% , c% = mdx100/ms
=> md = ms1xc%1/100 = 200x49/100 = 98 g
=>ms2 = mdx100/c%2 = 98x100/26,203 = 374 g
=> ms2 = ms1 + m.apa => m.apa = ms2 - ms1
=> m.apa = 374 - 200 = 174 g apa adaugata la diluarea sol. de H2SO4
m g ........................................................................ 174 g
CH3-CH2-CH2-OH _H+/toC_> CH3-CH=CH2 + H2O
60 g/mol .................................................................... 18 g/mol
=> m = 60x174/18 = 580 g propanol
verifica valoarea raspunsului..... !!!