Question

The freezing point of benzene is 5.5°C. What is the freezing point of a solution of 2.60 g of naphthalene (C10H8) in 675 g of benzene (Kf of benzene = 4.90°C/m)?

Answer 1

Answer: The freezing point of solution is 5.35°C

Step-by-step explanation:

The equation used to calculate depression in freezing point follows:

`\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}`

To calculate the depression in freezing point, we use the equation:

`\Delta T_f=iK_fm`

Or,

`\text{Freezing point of pure solution}-\text{Freezing point of solution}=i* K_f* \frac{m_(solute)* 1000}{M_(solute)* W_(solvent)\text{ (in grams)}}`

where,

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

`K_f` = molal freezing point elevation constant = 4.90°C/m

`m_(solute)` = Given mass of solute (naphthalene) = 2.60 g

`M_(solute)` = Molar mass of solute (naphthalene) = 128.2 g/mol

`W_(solvent)` = Mass of solvent (benzene) = 675 g

Putting values in above equation, we get:

`5.5-\text{Freezing point of solution}=1* 4.90^oC/m* (2.60* 1000)/(128.2g/mol* 675)\\\\\text{Freezing point of solution}=5.35^oC`

Hence, the freezing point of solution is 5.35°C