Answer: 8.42 grams of testosterone
Explanation:
Depression in freezing point is given by:
[tex]\Delta T_f=i\times K_f\times m[/tex]
[tex]\Delta T_f=0.500^0C[/tex] = Depression in freezing point
i= vant hoff factor = 1 (for non electrolyte)
[tex]K_f[/tex] = freezing point constant of benzene= [tex]5.12^0C/m[/tex]
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 (benzene)= 299.0 g = 0.299 kg
Molar mass of solute testosterone= 288.4 g/mol
Mass of solute testosterone added = ?
[tex]0.500=1\times 5.12\times \frac{xg}{288.4 g/mol\times 0.299kg}[/tex]
[tex]x=8.42g[/tex]
Thus 8.42 grams of testosterone must be dissolved in 299.0 grams of benzene to reduce the freezing point by 0.500°C.
