Question

Calculate the freezing point of a solution of 40.0 g methyl salicylate, c7h6o2, dissolved in 800. g of benzene, c6h6. and the freezing point is 5.50°c for benzene. calculate the freezing point of a solution of 40.0 g methyl salicylate, c7h6o2, dissolved in 800. g of benzene, c6h6. and the freezing point is 5.50°c for benzene. 3.41°c -2.09°c 7.59°c 2.09°c

Answer 1

From the problem statement, we are given a solution thus the solute in the solution would have an effect on some of the properties of the whole system. These properties are called the colligative properties. To calculate the freezing point of the solution, we use the freezing point depression equation which is expressed as follows:

ΔTf = kf(m)i

where ΔTf represents the freezing point depression, kf is a constant which 4.90 C/m for benzene, i is the vant hoff factor which is 1 for the given solute since it does not dissociate into ions and m is the molality of the solution. We calculate as follows:

ΔTf = kf(m)i
ΔTf = 4.90 (40.00 / .800 (122.13)) (1)
ΔTf = 2.01 C

ΔTf = Tf - Tfs
Tfs = 5.5 - 2.01
Tfs = 3.49 C

The correct answer would be the first option.

Answer 2

The freezing point of the solution will be 3.9075 $\rm ^\circ C$.

The freezing point of the solution will be calculated by the formula:

$\rm \Delta T_f\;=\;k_f\;\times\;molality\;\times\;i$

$\rm k_f$ is the constant = 4.90 C/m (benzene), i = von't hoff factor = 1

molality = $\rm \dfrac{weight}{molecular\;weight}\;\times\;mass\;of\;solution$

molality = $\rm \dfrac{40}{152}\;\times\;\dfrac{1000}{800}$

molality = 0.325 m

$\rm \Delta T_f$ = 4.90 $\times$ 0.325

$\rm \Delta T_f$ = 1.5925 $\rm ^\circ C$

The temperature of benzene is $\rm 5.50^\circ C$ and the change in temperature is 1.5925 $\rm ^\circ C$.

So, the solution temperature will be :

= 5.50 - 1.5925 $\rm ^\circ C$.

= 3.9075 $\rm ^\circ C$.

The freezing point of the solution will be 3.9075 $\rm ^\circ C$.

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