Hola

Solo disponible en BuenasTareas
  • Páginas : 3 (725 palabras )
  • Descarga(s) : 0
  • Publicado : 28 de abril de 2010
Leer documento completo
Vista previa del texto
Sustituyendo
M=0.324 gr0.082atm∙ltK∙mol300°K(0.734 atm)(1x10-3 lt)1+9(0.734 atm)(532.6 °K)128(39.48 atm)(300°K)1-6(532.6 °K)2(300°K)2
M=10.406x103grmol
Cuestionario
Anote sus resultadosexperimentales obtenidos:
Considerando comportamiento ideal, calcule el peso molecular de la sustancia problema:
PV = mMRT
P = 585 mmHg - P vapor del agua
P = 585 mmHg-26.8=558.21 atm760 mmHg=0.734 atm
T = 27 °C+273 °K=300 °K
M=0.324 gr0.082atm∙ltK∙mol300°K(0.734 atm)(1x10-3 lt)=10.858x103grmol
P = 585 mmHg -28.3=556.71 atm760 mmHg=0.732 atm
T = 28 °C+273°K=301 °K
M=0.324 gr0.082atm∙ltK∙mol301°K0.732 atm1x10-3 lt=10.924x103grmol
P = 585 mmHg -30.1=554.91 atm760 mmHg=0.730 atm
T = 29 °C+273 °K=302 °K
M=0.324 gr0.082atm∙ltK∙mol302°K0.730atm1x10-3 lt=10.991x103grmol
P = 585 mmHg -31.8=553.21 atm760 mmHg=0.727 atm
T = 30 °C+273 °K=303 °K
M=0.324 gr0.082atm∙ltK∙mol303°K0.727 atm1x10-3 lt=10.997x103grmol
P = 585 mmHg -33.7=551.31atm760 mmHg=0.725 atm
T =31 °C+273 °K=304 °K
M=0.324 gr0.082atm∙ltK∙mol304°K0.725 atm1x10-3 lt=11.033x103grmol
P = 585 mmHg -35.7=549.31 atm760 mmHg=0.722 atm
T = 32 °C+273 °K=305 °KM=0.324 gr0.082atm∙ltK∙mol305°K0.722 atm1x10-3 lt=11.223x103grmol
P = 585 mmHg -37.7=547.31 atm760 mmHg=0.720 atm
T =33 °C+273 °K=306 °K
M=0.324 gr0.082atm∙ltK∙mol306°K0.720 atm1x10-3lt=11.291x103grmol
P = 585 mmHg -39.9=545.11 atm760 mmHg=0.717 atm
T = 34 °C+273 °K=307 °K
M=0.324 gr0.082atm∙ltK∙mol307°K0.717 atm1x10-3 lt=11.375x103grmol
A partir de los pesos atómicosdetermine el peso molecular de la sustanciaproblema. Tenemos es reactivo CCl4
Donde C (carbono)= 12.01
Y Cl (Cloro) =35.42 (4)=141.68
MCCl4=MC+MCL4
MCCl4=12.01+141.68=153.69 grmolCalcule el peso molecular con la ecuación de Berthelot. *CCI4*: Tc = 532.6 °K Pc = 39.48 atm.
Sustituyendo
M=0.324 gr0.082atm∙ltK∙mol300°K(0.734 atm)(1x10-3 lt)1+9(0.734 atm)(532.6...
tracking img