Question

.ma poate ajuta cnv la astea?1 L solutie de H2So4 98% si p=1,84 g/cm3 se dilueaza cu 2 L de apa. a) calculati molaritatea solutiei initiale de H2So4 , b) calculati concentratia procentuala si concentratia molara  a  solutiei finale de H2So4...

Answer 1

Conform problemei :
Vs H2SO4 = 1L
densitate = 1.8 g / cm3
c = 98 %
V H20 = 2  L
a) densitate = ms / Vs -------->ms = densitate x Vs = 1.84 x 1L = 1.84 g 
                                            ms = 1.84 kg solutie H2SO4
md H2SO4 =c x ms / 100 = 98 x 1.84 / 100 = 1.80 kg
md H2SO4 = 1.80 kg H2SO4

n = numar de moli
M = masa moleculara
n H2SO4 = md H2SO4 / M H2SO4 = 1.80 / 98 = 0.01 kmoli
n H2SO4 = 0.01 kmoli

CM = concentratie molara
CM = n H2SO4 / Vs H2SO4 = 0.01 / 1 L = 0.01 M
CM H2SO4 = 0.01 M

b) densitate H2O = ms / Vs H2O ------->ms H2O = 1 g / cm3 x 2L = 2 kg 
                                                         ms H2O = 2 kg 
ms finala = ms H2SO4  + ms H2O = 1.84 + 2 kg = 3.84 kg solutie
msfinala = 1.84 kg solutie 
md = 1.80 kg  ------> c = md x 100 / ms = 1.8 x 100 / 3.84 = 46.87 %
c = 46.85 %

Vs final = 1 L + 2 L = 3 L
n = 0.01 kmoli
CM = n / Vs = 0.01 / 3 = 0.003 M