Question

amestecand 200 de grame de solutie de acid clorhidric de concentratie 20% cu o solutie de concentratie 36%, rezulta o solutie de concentratie 30%.calculati cate grame de solutie de concentratie 36% sunt necesare?

Answer 1

solutie 1 : ms1 = 200g    c1 = 20%  ⇒  md1 = 40g
solutie 2 : ms2                c2 = 36% ⇒  md2 = 0,36ms2
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solutie rezultata : ms = ms1 + ms2 = 200 + ms2 
md = md1 + md2 = 40 + 0,36ms2      c = 30%
(40+0,36ms2)·100/(200+ ms2) = 30
400 + 3,6ms2 = 600+ 3ms2
0,6ms2 = 200    ms2 = 333,33 g

Answer 2

1) Aflam masa dizolvata in prima solutie:

$c= \frac{md*100}{ms}$ ⇒ $md= \frac{c*ms}{100}$
In cazul nostru:

$md1= \frac{20*200}{100}=20*2=40g$

2) Exprimam si masa dizolvata in a doua solutie (cea de concentratie 36%)

$md2= \frac{c2*ms2}{100}= \frac{36*ms2}{100}=0,36*ms 2$

3) Aflam masa solutiei 2 folosindu-ne de solutia finala

Dupa amestecarea solutiilor vom avea:

msf=ms1+ms2=200+ms2 g (masa solutiei finale)
mdf=md1+md2=40+0,36*ms2 g (masa dizolvata in solutia finala)

Aplicam formula concentratiei si avem:

$cf= \frac{mdf*100}{msf}= \frac{(40+0,36ms2)*100}{200+ms2}= \frac{4000+36ms2}{200+ms2} =30$

Inseamna ca:

4000+36*ms2=30(200+ms2)
4000+36*ms2=6000+30*ms2
36*ms2-30*ms2=6000-4000
6*ms2=2000
ms2=333,33 g solutie

Deci raspunsul va fi 333,33 g solutie de concentratie 36%.