Question

20 puncteee
1.Care este concentratia molara a unei solutii de acid sulfuric de concentratie 18% cu ro=1,125 g?
2.Ce concentratie molara are solutia obtinuta prin amestecarea a 450 ml solutie NaOH 2M cu 150 ml solutie NaOH 6M?
3.Care este volumul solutiei de H2So4 2,5M care se poate obtine din 300g solutie H2SO4 98%?

Answer 1

1. c=18%
c=md/ms*100=> md/ms=18/100
md= 18/100*ms
1 mol H2SO4....98 g
n moli H2SO4...18*ms/100 g
n= (18*ms/100):98= 18*ms/9800
ro=ms/Vs
Vs= ms/ro= ms/1,125
cM= n/Vs= (18*ms/9800):(ms/1,125)= 18*1,125/98= 0,206 M
2.Vs1=450 ml= 0,45 l
cM1= 2
n1= cM1*Vs1=2*0,45= 0,9 moli NaOH
Vs2=150 ml= 0,15 l
cM2= 6
n2= cM2*Vs2= 6*0,15=0,9 moli NaOH
n final= n1+n2= 1,8 moli NaOH
Vs final= Vs1+Vs2=0,6 l
cM final= n final/Vs final= 1,8/0,6= 3 M
3.ms= 300 g H2SO4
c= 98%
md= ms*c/100= 294 g H2SO4
1 mol H2SO4...98 g H2SO4
n moli H2SO4...294 g H2SO4
n= 294/98= 3 moli
cM= 2,5 M
Vs=n/cM=3/2,5=1,2 l solutie H2SO4