Question

intr-un vas se gaesc 832,5g de solutie saturata de AgNO3 , la temperatura de 50frade C. Seraceste slutia la 20 gradeC. Calculati masa de AgNO3 depusa la racire, stiind ca solutia azotatului de argint e 455g la 50 gradeC si 222 g la 20 grade C.

Answer 1

50 grade se gasesc....455 g AgNO3 = md1............455+100=555 g AgNO3 = ms1
c1%=md/ms*100=455/555*100=81,98%
20 grade se gasesc 222 g AgNO3 = md2..........222+100=322 g AgNO3 = ms2
c2=md/ms*100=222/322=69,94%
md la 50 grade = c1 * ms total/100 = 81,98*832,5/100=682,48g
md la 20 grade = c2 * ms total/100= 69,94*832,5/100=582,25g
m apa la 50 grade = 832,5 - 682,48 = 150,02 g apa
la 20 grade:100 g apa.........222 g AgNO3
                 150,02 g AgNO3........x           x = 333,04 g AgNO3
m depusa = 682,48 - 333,04 = 349,44 g azotat