Question

Printr-un litru de apa a fost barbotat metanal cu volumul de 11,2 l.Care este masa solutie formate?

Answer 1

n=V/Vm=11,2/22,4=0,5 moli CH2O

md=30*0,5=15g CH2O

mH2O=1000g

ms = md+mH2O=15+1000=1015g sol.

n(niu) - nr de moli, mol

V-volumul,L

Vm- volumul molar (volumul unui gaz in c.n) = 22,4L

M-masa moleculara, g/mol

.Am aplicat relatia n=V/Vm= 11,2/22,4 =0,5 moli aldehida formica (metanal) CH2O

MCH2O=AC+2AH+AO=12+2*1+16=30g/mol

n=md/M=>md=n*M=0,5*30 sau 1/2*30 = 30/2 = 15 g CH2O

roH2O=1g/cm3

1L=1000ml sau 1000cm3

1mL=1cm3

VH2O=1L=1000cm3

roH2O=mH2O/V => mH2O=roH2O*VH2O=1g/cm3*1000cm3 = 1000g (am simplificat cm3 cu cm3)

ms = md+mH2O=15+1000=1015g sol.

Sper ca intelegi acum..