Question

Ethylene glycol, C2H6O2, is a nonvolatile substance unable to form ions in water. If 38.6 grams of ethylene glycol is dissolved in 175 grams of water, what is the freezing point of the solution? Kf = 1.86°C/m; Kb = 0.512°C/m

-6.61°C

1.82°C

6.61°C

-1.82°C

Answer 1

We gotta use the formula for f.p. depression.

\(\sf \huge \Delta T=i*m*K_f\)

\(\sf \Delta T \) is the change in the temperature from the normal value
\(\sf i\) is the van't hoff constant, it's 1 in this question
\(\sf m\) is molality
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it's defined as:

\(\sf molality =\dfrac{moles~solute}{Kg~solvent}\)
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\(\sf K_f\) is the f.p. depression constant (given in the question)


There are several steps to do this question. The first is converting 38.6 grams of ethylene glycol to moles.

We will use the definition of moles: \(\sf moles=\dfrac{mass}{Molar~mass}\)

\(\sf moles=\dfrac{38.6~g}{Molar~mass}\)

We also need to calculate the molar mass.
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The molar mass is the sum of the molar masses (found on the periodic table) according to the formula for ethylene glycol.

\(C_2H_6O_2\)

carbon - 12 g/mol
hydrogen - 1 g/mol
oxygen - 16 g/mol

\(\sf Molar~mass~of~C_2H_6O_2 =(12~g/mol*2)+(1~g/mol*6)+(16~g/mol*2)=62~g/mol\)

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\(\sf moles=\dfrac{38.6~g}{62~g/mol}=0.623~moles \)

Now we find the molality:

\(\sf molality =\dfrac{0.623~moles}{0.175 kg}=3.56 ~m\)

(Note:converted grams of water to kg)

Finally, we can calculate the change in the temperature (from the normal value)

\(\sf \Large \Delta T=i*m*K_f\)

\(\sf \Large \Delta T=1*3.56 ~m*(-1.86°C/m)\)