Question

Se amesteca 2 kg solutie de sare 35% cu 5 kg solutie de sare 15% si 3 kg de apa .Aflati concentratia solutiei finale

Answer 1

$md_{1}$ = $\frac{ms*c}{100} = \frac{2*35}{100} = \frac{70}{100} = 0,7$ kg sare

$md_{2}$ = $\frac{5*15}{100} = \frac{75}{100} = 0,75$ kg sare

$ms_{final} = 2 + 5 + 3 = 10 kg$

$md_{final} = 0,7 + 0,75 = 1,45$ kg sare

$c_{finala} = \frac{ md_{final}}{ms_{final}} * 100 = \frac{1,45*100}{10}$ = 14,5%

Answer 2

ms1=2kg=2000g
C1=35%
md1=$\frac{C1*ms1}{100}$= $\frac{35*2000}{100}$= 700 g
ms2=5 kg=5000g
C2=15%
md2= $\frac{C2*ms2}{100}= \frac{15*5000}{100}$= 750g
mH2O= 3kg= 3000g
mdfinal= md1+md2= 700+750=1450g
msfinal= ms1+ms2+mH2O= 2000+5000+3000= 10000g
Cfinal= $\frac{mdfinal}{msfinal}*100$= $\frac{1450}{10000}*100$= 14,5 %