Respuesta :
Answer:
A)Boiling point constant of benzene = 2.63°C/m
B) 242.77 g/mol is the molar mass of the solute.
Explanation:
[tex]\Delta T_b=T_b-T[/tex]
[tex]\Delta T_b=K_b\times m[/tex]
[tex]Delta T_b=iK_b\times \frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in Kg}}[/tex]
where,
[tex]\Delta T_f[/tex] =Elevation in boiling point
[tex]K_b[/tex] = boiling point constant od solvent= 3.63 °C/m
1 - van't Hoff factor
m = molality
A) Mas of solvent = 500 g = 0.500 kg
T = 80.1°C ,[tex]T_b[/tex] =82.73°C
[tex]\Delta T_b=T_b-T[/tex]
[tex]\Delta T_b[/tex]= 82.73°C - 80.1°C = 2.63°C
[tex]2.63^oC=K_b\times \frac{36 g}{72 g/mol\times 0.500 kg}[/tex]
[tex]K_b=\frac{2.63^oC\times 72 g/mol\times 0.500 kg}{36 g}=2.63 ^oC/m[/tex]
Boiling point constant of benzene = 2.63°C/m
B) Mass of solute = 1.2 g
Molar mas of solute = M
Mass of solvent = 50 g= 0.050 kg
i = 1
T = 80.1°C ,[tex]T_b[/tex] =80.36°C
[tex]\Delta T_b=T_b-T[/tex]=80.36°C - 80.1°C = 0.26°C
[tex]0.26^oC=1\times K_b\times \frac{1.2 g}{M\times 0.050 kg}[/tex]
[tex]M=1\times 2.63^oC/m\times \frac{1.2 g}{0.26^oC\times 0.050 kg}[/tex]
M = 242.77 g/mol
242.77 g/mol is the molar mass of the solute.
