Question

Calculate the mass of ascorbic acid (${{C}_{6}}{{H}_{8}}{{O}_{6}}$) to be dissolved in 75 g of acetic acid to lower the melting by ${{1.5}^{\circ }}C$ (${{K}_{f}}$for acetic acid is 3.9 K kg/mol).

Answer 1

Hint: The mass of ascorbic acid can be calculated by the formula of depression in freezing point, \[\Delta {{T}_{f}}={{K}_{f}}\text{ x }\dfrac{{{w}_{2}}}{{{M}_{2}}}\text{ x }\dfrac{1000}{{{w}_{1}}}\], \[\Delta {{T}_{f}}\] is the depression in freezing point, \[{{w}_{2}}\] is the mass of the solute in gram, \[{{M}_{2}}\] is the molar mass of the solute, and\[{{w}_{1}}\] is mass of the solvent in gram.

Complete step by step answer:
Let us understand about the depression in freezing point:
The freezing point of a substance is the temperature at which the solid and the liquid form of the substance are in equilibrium i.e., the solid and liquid form has the same vapour pressure. It is noticed that a solution’s freezing point is always lower than the freezing point of the pure solvent present in the solution. This is called the depression in freezing point.
The depression of freezing point depends on the solute in a solution and has found to be related to the molality’ as below:
\[\Delta {{T}_{f}}={{K}_{f}}\text{ x }m\]
\[{{K}_{f}}\]is the molal depression constant.
So, by expanding the above formula, we get:
\[\Delta {{T}_{f}}={{K}_{f}}\text{ x }\dfrac{{{w}_{2}}}{{{M}_{2}}}\text{ x }\dfrac{1000}{{{w}_{1}}}\]
Where, \[\Delta {{T}_{f}}\]is the depression in freezing point,
\[{{w}_{2}}\] is the mass of the solute in gram,
\[{{M}_{2}}\] is the molar mass of the solute,
\[{{w}_{1}}\] is the mass of the solvent in grams.
So, according to the question:
\[{{w}_{2}}\] is the mass of solute (ascorbic acid) = x
\[{{M}_{2}}\]= Molar mass of the solute (ascorbic acid, ${{C}_{6}}{{H}_{8}}{{O}_{6}}$) = $6\text{ x 12 + 1 x 8 + 6 x 16 = 176 g/mol}$
\[{{w}_{1}}\]= Mass of the solvent (acetic acid) = 75 g
\[\Delta {{T}_{f}}={{1.5}^{\circ }}C\]
${{K}_{f}}$for acetic acid = 3.9 K kg/mol
Putting all these in the formula, we get
\[\Delta {{T}_{f}}={{K}_{f}}\text{ x }\dfrac{{{w}_{2}}}{{{M}_{2}}}\text{ x }\dfrac{1000}{{{w}_{1}}}\]
\[{{w}_{2}}=\text{ }\dfrac{\Delta {{T}_{f}}}{1000}\text{ x }\dfrac{{{M}_{2}}}{{{K}_{f}}}\text{ x }{{w}_{1}}\]
\[{{w}_{2}}=\text{ }\dfrac{1.5}{1000}\text{ x }\dfrac{176}{3.9}\text{ x 75 = 5}\text{.08 g}\]

So, the mass of ascorbic acid required to dissolve is 5.08 g.

Note: The mass of the solvent should always be taken in grams, if it is given in litres, then convert it into grams. \[{{K}_{f}}\]is also called the cryoscopic constant of the solvent. Depression in freezing point is a colligative property because it depends on the number of moles of the solute.