contestada

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 [tex]K_f[/tex] 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 [tex]K_f[/tex] that you calculated in question 2.

Respuesta :

Answer:

The value of [tex]K_f[/tex] for xylene is 4.309°C/m.

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

Explanation:

[tex]\Delta T_f=K_f\times \frac{\text{Amount of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent(kg)}}[/tex]

where,

[tex]\Delta T_f[/tex] =depression in freezing point

[tex]K_f[/tex] = freezing point constant  

we have :

1) freezing point constant  for xylene = [tex]K_f[/tex] =?

Mass of toluene = 0.193 g

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

[tex]\Delta T_f=3.57^oC[/tex]

[tex]3.57^oC=K_f\times \frac{0.193 g}{92 g/mol\times 0.002532 kg}[/tex]

[tex]K_f=4.309^oC/m[/tex]

The value of [tex]K_f[/tex] 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 = [tex]K_f=4.309^oC/m[/tex]

[tex]2.88^oC=4.309^oC/m\times \frac{0.123g}{M\times 0.002493 kg}[/tex]

M = 73.82 g/mol

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

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