Question

In 72 g solutie 11,11% Na OH se introduc 8gNaOH.Calculati concentratia procentuala si concentratia molara a solutiei dinale,daca densitatea acesteia este:1.225g/mL

Answer 1

c=(md/ms)*100 => md=(c*ms)/100=(11,11*72)/100=7,9992=8
md2=8+8=16 g 
ms2=72+8=80 g => c2=(md2/ms2)*100=(16/80)*100=20%
C=Nr. moli/Vs
d=ms/Vs
Nr. moli=md2/masa molara =>Nr. moli=16/40=0,4 moli
masa molaraNaOH=40 g/mol
Vs=ms/d=80/1,225=65,3061 ml=0,065 L
=>C=0,4/0,065=6,15 M