Question

1) If 0.193 grams of toluene is dissolved in 2.532 grams of p-xylene, what is the molality of toluene in the solution?2) If a freezing point depression of 3.57°Celcius is measured for the solution described in question 1, calculate
`K_f` for p-xylene.3) Suppose you dissolved 0.123 gram of pentane in 2.493 grams of p-xylene and measured a freezing point depression of 2.88°celcius for the solution. Calculate the molar mass of pentane using this data and the value for
`K_f` that you calculated in question 2.

Answer 1

The value of
`K_f` for xylene is 4.309°C/m.

The molar mass of pentane using this data is 73.82 g/mol.

Step-by-step explanation:

`\Delta T_f=K_f* \frac{\text{Amount of solute}}{\text{Molar mass of solute}* \text{Mass of solvent(kg)}}`

where,

`\Delta T_f` =depression in freezing point

`K_f` = freezing point constant

we have :

1) freezing point constant for xylene =
`K_f` =?

Mass of toluene = 0.193 g

Mass of xylene = 2.532 kg = 0.002532 kg ( 1 g =0.001 kg)

`\Delta T_f=3.57^oC`

`3.57^oC=K_f* (0.193 g)/(92 g/mol* 0.002532 kg)`

`K_f=4.309^oC/m`

The value of
`K_f` for xylene is 4.309°C/m.

2)

Mass of pentane = 0.123 g

molar mass of pentame= M

Mass of xylene = 2.493 g = 0.002493 kg

Freezing point Constant of xylene =
`K_f=4.309^oC/m`

`2.88^oC=4.309^oC/m* (0.123g)/(M* 0.002493 kg)`

M = 73.82 g/mol

The molar mass of pentane using this data is 73.82 g/mol.