Question

Calculate the freezing point of a solution made from 220g of octane (C Hua), molar mass = 114,0 gmol dissolved in 1480 g of benzene. Benzene freezes at 5.50"C and its Kvalue is 5.12C/m. -1.16°C 0.98°C 666"C 12 2°C 5.49°C 10 12 AM A A 2019 Backspace yuo Pill но кL

Answer 1

Answer: Freezing point of a solution will be $-1.16^0C$

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=(5.50-T_f)^0C$ = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

$K_f$ = freezing point constant = $5.12^0C/m$

m= molality

$\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}$

Weight of solvent (benzene)= 1480 g =1.48 kg

Molar mass of solute (octane) = 114.0 g/mol

Mass of solute (octane) = 220 g

$(5.50-T_f)^0C=1\times 5.12\times \frac{220g}{114.0 g/mol\times 1.48kg}$

$(5.50-T_f)^0C=6.68$

$T_f=-1.16^0C$

Thus the freezing point of a solution will be $-1.16^0C$