Question

Una disoluci´on de 2.50 g de un compuesto de f´ormula emp´ırica C6H5P en 25.0 g de benceno se
congela a 4.3◦C. Calcular la masa molar y la f´ormula molecular del soluto teniendo en cuenta
que el benceno puro se congela a 5.5◦C. Kf (benceno) = 5.12◦C kg mol−1

Answer 1

Δtc = Kc · m      (m = molalidad)

m = mol sto / Kg svte.

m = masa soluto  (Fórmula 1)
       ````````````````
       Mm · Kg svte

1. calcular Δtc = 4.3 ºC - 5.5 ºC = - 1.2 ºC

Tc = 0º C - Δtc
Tc = 0ºC -1.2)ºC = 1.2 ºC

2. Calcular Mm

ΔTc = Kc · m  sustituir m por fórmula 1

Δtc = Kc . masa soluto
                 ``````````````````                  despejar Mm
                  Mm · Kg svte.

Mm =  Kc · masa soluto
           `````````````````````````
               Δtc · Kg svte

svte = 25 g de benceno /1000 = 0.025 Kg

Mm = 5.12 ºC Kg /mol  · 2.50 g
         ``````````````````````````````````````
             0.025 Kg · 1.2 ºC

Mm = 426.66 g/mol

b) calcular fórmula molecular de C6H5P
Mm del C6H5P
C: 6 x 12 = 72 g/mol
H= 5 x 1 = 5 g/mol
P = 1 x 30.9 = 30.9 g/mol
````````````````````````````````````
         Mm = 107.9 g/mol
calcular n

n = 426.66 g/mol
     `````````````````````
       107.9 g/mol

n = 4

FM: (C6H5P)4 = C24H20P4