Question

1) 5,6 ml (c.n.) propena pura se barboteaza in 250 ml solutie de brom in apa pana la decolorarea completa a acesteia. Calculeaza concentratia molara a solutiei de brom.

2) In urma analizei a 40 mg alchină se obtine un volum de 67,2 ml CO2 ( cn ). Determina prin calcul formula moleculara a alchinei stiind ca prin aditia HBr la alchina determinata se obtine un compus care nu contine carbon tertiar .

Answer 1

n,propena = 0,0056l/22,4l/mol=0,00025 mol
CH2=CH-CH3+Br2---> CH2Br-CHBr-CH3
 1mol................1mol
 0,00025mol......0,00025mol
c=n/V;   c=0,00025mol/ 0,250l=0,001mol/l


alchina are formula generala CnH2n-2   si M= 12n-2 g/mol
V,CO2= 67,2ml---> n= 0,0672 l/22,4l/mol=0,003molC02---> 0,003molC
m,C= 0,003MOLX12g/mol= 0,036g C
0,040g alchina............0,036gC.....
14n-2g........................12n...
0,48n= 0,504n-0,072---> n=3---> propina

SE  putea lucra cu cifre fara asa multe zero-uri daca luam mg, ml,mmol si la sfarsit transformam in mol !!