Question

What is the freezing point of water made by dissolving 22.78 g of ethylene glycol (CH2(OH)CH2(OH)) in 87.95 g of water? The freezing-point depression constant of water is 1.86 oC/m.

Answer 1

Explanation:

When a non-volatile solute is added in a solvent then decrease in its freezing point is known as freezing point depression.

Mathematically,     $\Delta T = K_{f}m$

where      $\Delta T$ = change in freezing point

                     $K_{f}$ = freezing point depression constant $\text({in ^{o}C/mol/kg})$

                           m = molality

First, calculate the number of moles as follows.

                       No. of moles = $\frac{\text{mass of solute}}{\text{molar mass of solute}}$

                                             = $\frac{22.78g}{62.07 g/mol}$

                                             = 0.367 mol

Now, it is given that mass of solvent is 87.95 g. As there are 1000 grams in 1 kg.

So,          $87.95g \times \frac{1 kg}{1000 g}$ = 0.08795 kg

Hence, molality of the given solution is as follows.

                   Molality = $\frac{\text{no. of moles}}{\text{mass in kg}}$

                                 = $\frac{0.367 mol}{0.08795}$

                                 = 4.172 mol/kg

Therefore, depression in freezing point will be as follows.

                          $\Delta T = K_{f}m$

            $T_{solvent} - T_{mixture}$ = $K_{f}m$

Since, freezing point of pure water is $0^{o}C$. Now, putting the given values as follows.

                $0^{o}C - T_{mixture}$ = $1.86^{o}C/m \times 4.172mol/kg$

                                         = $-7.759 ^{o}C$

Thus, we can conclude that the freezing point of water in the given mixture is $-7.759 ^{o}C$.