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Calculate the freezing point of a solution made from 52.6 g of propane, c3H8, dissolved in 196.0 g of benzene, C6H6. The freezing point of benzene is 5.50 C and its Kf is 5.12 C/m.

Respuesta :

PBCHEM
Given: weight of solute (propane) = 52.6 g
weight of solvent (benzene) = 196 g = 0.196 kg
We know that, molecular weight of propane = 44.1 g/mol

∴ Molality of solution = [tex] \frac{\text{weight of solute (g)}}{\text{Molecular weight X Weight of solvent (Kg)}} [/tex]
                                  =  [tex] \frac{52.6}{44.1X.196} [/tex] = 6.085 m

Now, Depression of freezing point = Kf m
where Kf = cryoscopic constant = 5.12 oC/m

∴  Depression of freezing point = 5.12 X 6.085
                                                  = 31.15 oC

Therefore, freezing point of solution = 5.50 - 31.15 = -25.65 oC

The freezing point is the temperature at which the substance freezes. The freezing point of the solution is -25.65 degrees celsius.

What is the depression of freezing point?

The depression of the freezing point is the colligative property of the solutions and is due to the addition of the solute in the solvents.

Given,

Mass of the solute = 52.6 gm

Mass of the solvent = 0.196 gm

Molecular mass of solute = 44.1 g/mol

Substituting values in the equation of molality:

[tex]\begin{aligned}\rm Molality &= \rm \dfrac{\text{weight of solute}}{\text{weight of solvent }\times \text{molecular mass solute}}\\\\&=\dfrac{52.6}{44.1 \times 0.196}\\\\&= 6.085\;\rm m\end{aligned}[/tex]

Now, the freezing point depression can be estimated as:

[tex]\begin{aligned}\text{Depression of freezing point} &= \rm K_{f} \times m\\\\&= 5.12 \times 6.085\\\\&= 31.15^{\circ}\;\rm C\end{aligned}[/tex]

The freezing point of the solution will be, 5.50 - 31.15 = -25.65 degrees celsius.

Therefore, -25.65 degrees celsius is the freezing point of the solution.

Learn more about a freezing point here:

https://brainly.com/question/4475402

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