Question

Calculate the mass of ascorbic acid \((\mathrm{C_6H_8O_6})\) to be dissolved in \(75\) g of acetic acid to lower its melting point by \(1.5^\circ \mathrm{C}\). \(K_f = 3.9\ \mathrm{K\ kg\ mol^{-1}}\); \(H = 1,\ C = 12,\ O = 16\ amu\)

• A. 5.770 g
• B. 4.077 g
• C. 5.077 g
• D. 4.770 g

Hint:

In freezing point depression problems, first calculate molality using \( \Delta T_f = K_f m \), then find moles of solute from the mass of solvent in kilograms, and finally convert moles into mass using molar mass.

Answer 1

The Correct Option is C

Step 1: Use the formula for depression in freezing point.
The depression in melting point or freezing point is given by
\[ \Delta T_f = K_f \times m \] where \(m\) is the molality of the solution. Therefore,
\[ m = \frac{\Delta T_f}{K_f} = \frac{1.5}{3.9} \] \[ m = 0.3846\ \mathrm{mol\ kg^{-1}} \] Step 2: Convert mass of solvent into kilogram.
Mass of acetic acid \(= 75\ \mathrm{g} = 0.075\ \mathrm{kg}\)
Now, molality is defined as
\[ m = \frac{\text{moles of solute}}{\text{kg of solvent}} \] So, moles of ascorbic acid required are
\[ \text{moles of solute} = m \times \text{kg of solvent} \] \[ = 0.3846 \times 0.075 \] \[ = 0.028845\ \text{mol} \] Step 3: Calculate molar mass of ascorbic acid.
Ascorbic acid is \( \mathrm{C_6H_8O_6} \). Its molar mass is
\[ (6 \times 12) + (8 \times 1) + (6 \times 16) \] \[ = 72 + 8 + 96 = 176\ \mathrm{g\ mol^{-1}} \] Step 4: Calculate the required mass.
Mass of ascorbic acid \(=\) moles \(\times\) molar mass
\[ = 0.028845 \times 176 \] \[ = 5.07672\ \mathrm{g} \] \[ \approx 5.077\ \mathrm{g} \] Step 5: Conclusion.
Hence, the mass of ascorbic acid required to produce the given lowering in melting point is 5.077 g.
Final Answer:5.077 g.