Quimica

TETRACLORURO DE CARBONO (CCl4)

*Cálculos:

M muestra 0.015 gr.

T 28 °C

V desplazado 0.0033L

1. Considerando comportamiento ideal, calcule el peso molecular de la sustancia problema.

*Cálculos:

mmuestra: 2.040gr. – 2.025gr. = 0.015gr.

Volumen:

(3.3ml)*((1L)/(1000ml))= 0.0033L.

Presion:

585 mmHg – 28.3mm Hg = 556.7 mm Hg

((556.7mm Hg)*(1atm)/(760mm Hg)) = 0.7325 atm

Temperatura:

28°C+273.15 = 301.15 K

M:

M=((m*R*T)/(P*V))

M=(((0.015gr)*(0.08205atm*l/mol*K)(301.15K))/(0.7325atm)(0.0033L))

M=153.3314 gr./mol

2. A partir de los pesos atómicos determine el peso molecular de la sustancia problema

Peso C: 12gr.

Peso Cl: 35gr.

CCl4: (12gr*1)+(35gr*4)=152gr./mol

3. Calcule el peso molecular con la ecuación de Berthelot

PV=nRT {1+[((9PTC)(128PTC))*(1-(6TC²/T²))]}

Donde

n=(m/(PM))

entonces

PV=((m/(PM))*(RT)) {1+[((9PTC)/(128PCT))*(1-(6TC²/T²))]}

Por lo tanto

PV=((m*R*T/(PM)) {1+[((9PTC)/(128PCT))*(1-(6TC²/T²))]}

CCl4: Tc = 532.6 °K Pc = 39.48 atm.

M = (153.3314 gr./mol) {1+[((9PTC)/(128PCT))*(1-(6TC²/T²))]}

M = (153.3314 gr./mol) {1+[((9(0.7325 atm)( 532.6 °K))/((128(39.48 atm)(301.15 K)))*(1-(6(532.6 °K)²/(301.15K)²))]}

M = (153.3314 gr./mol) (.4590)

M = 147.04 gr/mol

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