-calculez conc.procentuala a solutiei molare, stiind M,NaCl= 58,5g/mol
3x58,5g NaCl.........1000ml..........1000x1,12g
c1.......................................................100g
c1= 15,67%
-deduc ms1 din ecuatia:
c2= md2x100/ms2 unde md2=md1= c1xms1/100 ms2=ms1+m,apa
10=(15,67xms1)/(ms1+200)
10ms1+2000=15,67ms1====> ms1=352,7g---> v,s1= 352,7/1,12ml=315ml