How many grams of boric acid do you need to prepare 300 mL of a solution with a freezing point of -0.64oC. Boric acid has an E value of 0.5.
The correct answer and explanation is :
To determine the grams of boric acid needed, we follow these steps:
Step 1: Use Freezing Point Depression Formula
[
\Delta T_f = i \cdot K_f \cdot m
]
where:
- (\Delta T_f) = Freezing point depression = (0 – (-0.64) = 0.64^\circ C)
- (i) = Van’t Hoff factor (for boric acid, it is 1 since it does not ionize significantly)
- (K_f) = Cryoscopic constant for water = (1.86^\circ C \cdot kg/mol)
- (m) = Molality (moles of solute per kg of solvent)
Step 2: Solve for Molality
[
m = \frac{\Delta T_f}{i \cdot K_f} = \frac{0.64}{1 \times 1.86} = 0.3441 \text{ mol/kg}
]
Step 3: Convert to Moles of Boric Acid
The molecular weight of boric acid (H₃BO₃) is 61.83 g/mol.
[
\text{Mass} = \text{Moles} \times \text{Molar Mass} = 0.3441 \times 61.83 = 21.28 \text{ g} \text{ per kg of solvent}
]
Step 4: Adjust for Volume
Since 300 mL (0.3 kg of water) is used:
[
\text{Mass} = 21.28 \times 0.3 = 6.38 \text{ g}
]
Step 5: Adjust for E Value
Boric acid has an E value of 0.5, meaning its actual effect on freezing point is only half of its weight.
[
\text{Required Mass} = \frac{6.38}{0.5} = 12.76 \text{ g}
]