Respuesta :
Answer:
The boiling point of a 2.47 m solution of naphthalene is 86.25°C.
The freezing point of a 2.47 m solution of naphthalene is -8.53°C.
Explanation:
Elevation in boiling point:
[tex]\Delta T_b=T_b-T[/tex]
[tex]\Delta T_b=K_b\times m[/tex]
where,
[tex]\Delta T_b[/tex] =elevation in boiling point =
[tex]K_b[/tex] = Boiling point constant of solvent
[tex]T_b,T[/tex] = boiling point of solution and pure solvent
m = molality
we have :
[tex]K_b[/tex] =2.53°C/m ,
m = 2.47m
[tex]\Delta T_b=2.53^oC\times 2.47m[/tex]
[tex]\Delta T_b=6.25^oC[/tex]
Boiling point of pure benzene= T = 80.1°C
Boiling point of solution = [tex]T_b[/tex]
[tex]\Delta T_b=T_b-T[/tex]
[tex]T_b=T+\Delta T_f=6.25^oC+80.1^oC=86.35^oC[/tex]
The boiling point of a 2.47 m solution of naphthalene is 86.25°C.
Depression in freezing point:
[tex]\Delta T_f=T-T_f[/tex]
[tex]\Delta T_f=K_f\times m[/tex]
where,
[tex]\Delta T_f[/tex] =depression in freezing point =
[tex]K_f[/tex] = freezing point constant of solvent
[tex]T_f,T[/tex] = freezing point of solution and pure solvent
m = molality
we have :
[tex]K_f[/tex] =5.12°C/m , 2.53
m = 2.47m
[tex]\Delta T_f=5.12^oC\times 2.47m[/tex]
[tex]\Delta T_f=14.03^oC[/tex]
Freezing point of pure benzene= T = 5.5°C
Freezing point of solution = [tex]T_f[/tex]
[tex]\Delta T_f=T-T_f[/tex]
[tex]T_f=T-\Delta T_f=5.5^oC-14.03^oC=-8.53^oC[/tex]
The freezing point of a 2.47 m solution of naphthalene is -8.53°C.
