Respuesta :
Answer: The freezing point of the new solution is [tex]4.3^0C[/tex]
Explanation:
Depression in freezing point :
[tex]\Delta T_f=i\times K_f\times m\\\\\Delta T_f=i\times K_f\times\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in kg}}[/tex]
Mass of solute (tetraphenylporphryin) = 37 g
Mass of benzene = 250 g = 0.25 kg (1 kg = 1000 g)
Formula used :
where,
[tex]\Delta T_f[/tex] = change in freezing point = [tex]T_f^0-T_f=5.5^0C-T_f[/tex]
i = Van't Hoff factor = 1 (for non electrolytes)
[tex]K_f[/tex] = freezing point constant = 5.12 K.kg/mole
m = molality
Now put all the given values in this formula, we get
[tex]5.5^0C-T_f=1\times (5.12K.kg/mole)\times \frac{37g}{614.7\times 0.25kg}[/tex]
[tex]5.5^0C-T_f=1.2[/tex]
[tex]T_f=4.3^0C[/tex]
Thus the freezing point of the new solution is [tex]4.3^0C[/tex]
