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`

Step-by-step explanation:

Depression in freezing point is given by:

`\Delta T_f=i* K_f* 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* K_f* \frac{\text{mass of solute}}{\text{molar mass of solute}* \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* 5.12* (220g)/(114.0 g/mol* 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`