Question

a)Solutia rezultata prin amestecarea a 250 ml solutie de NaOH cu p=1.437 g/cm3 si concentratia 40% cu 200 g H2O
b) solutia rezultata prin introducerea a 4.6 g Na in 200 g solutie NaOH de concentratie 4%
Va rog ajutati-ma!

Answer 1

a) Vs₁ = 250mL sol. NaOH
ρ₁ = 1,437g/cm³
C₁ = 40%
m H₂O adaugata = 200g

REZ:

ρ₁ = ms₁/Vs₁ ⇒ ms₁ = ρ₁×Vs₁ = 1,437×250 = 359,25g sol. NaOH

C₁ = md₁/ms₁×100 → md₁ = C₁×ms₁/100 = 40×359,25/100 = 143,7g NaOH

mdf = md₁ = 143,7g NaOH
msf = ms₁+mH₂O adaugata  = 359,25+200 = 559,25g sol. NaOH

Cf = mdf/msf×100 = 14370/559,25 = 25,7 %

b) C = md/ms×100 ⇒ md = C×ms/100 = 4×200/100 = 8g NaOH

ms = md+mH₂O ⇒ mH₂O = ms-md = 200-8 = 192g

n Na = 4,6/23 = 0,2 moli 

Na + H₂O ⇒ NaOH + ¹/₂H₂↑

1 mol Na............18g H₂O..............40g NaOH.........1g H₂
0,2 moli Na.........x............................y.......................z

x = 3,6g H₂O
y = 8g NaOH
z = 0,2g H₂

mdf = 8+8 = 16g NaOH

msf = 8+8+(192-3,6)-0,2 = 204,2g sol. NaOH

Cf = mdf/msf×100 = 16/204,2×100 = 7,835 %