Question

3.3 punctul c și 4.4 vă rog

Answer 1

3.3. a) Vs = 200mL
n = 20 mmoli = 0,02 moli NaOH

C = n/Vs(L) = 0,02/0,2 = 0,1M

b) Vs = 100mL
md = 8g NaOH

C = md/M×Vs(L) = 8/40×0,1 = 2M

c) ms = 200g sol. NaOH
C = 20%
ro = 1,225g/cm^3
VH2O = 36,74mL

C = md/ms×100 => md = C×ms/100 = 20×200/100 = 40g NaOH

ro = ms/Vs => Vs = ms/ro = 200/1,225 = 163,2653 mL = 0,1632653 L sol. NaOH

Vs final = V sol. + VH2O = 36,74+163,2653 = 200mL = 0,2 L

C = md/M×Vsf(L) = 40/40×0,2 = 5M

4.4. C = 98%
ro = 1,84g/cm^3
ms = 20g sol. H2SO4
Cfinal = 5M

C = md/ms×100 => md = C×ms/100 = 98×20/100 = 19,6g H2SO4

n = 19,6/98 = 0,2 moli

ro = ms/Vs => Vs = ms/ro = 20/1,84 = 10,87 mL sol. H2SO4

Cf = n/Vs => Vs = n/C = 0,2/5 = 0,04 L = 40mL sol. necesara

VH2O necesar diluarii = 40-10,87 = 29,13mL