Answer: The freezing-point depression constant (Kf) of nitrobenzene is [tex]5.7^0C/m[/tex]
Explanation:
Depression in freezing point is given by:
[tex]\Delta T_f=i\times K_f\times m[/tex]
[tex]\Delta T_f=T_f^0-T_f=(5.67-(-0.53)^0C=6.2^0C[/tex] = Depression in freezing point
i= vant hoff factor = 1 (for non electrolyte nitrobenzene)
[tex]K_f[/tex] = freezing point constant = ?
m= molality
[tex]\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]
Weight of solvent = 20 g = 0.02 kg
mass of solute (ethanol) = 1.0 g
Molar mass of ethanol = 46 g/mol
[tex]6.2=1\times K_f\times \frac{1.0g}{46g/mol\times 0.02kg}[/tex]
[tex]K_f=5.7^0C/m[/tex]
Thus freezing-point depression constant (Kf) of nitrobenzene is [tex]5.7^0C/m[/tex]
