Answer:
Freezing T° of solution = - 8.98°C
Explanation:
We apply Freezing point depression to solve this problem, the colligative property that has this formula:
Freezing T° of pure solvent - Freezing T° of solution = Kf . m
Kf = 1.86°C/m, this is a constant which is unique for each solvent. In this case, we are using water
m = molality (moles of solute / 1kg of solvent)
We convert the mass of solvent from g to kg
1.5 g . 1kg/1000g = 0.0015 kg
We convert the mass of solute, to moles. Firstly we make this conversion, from mg to g → 450mg . 1g/1000mg = 0.450 g
0.450 g. 1mol / 62.07g = 7.25×10⁻³ moles
Molality → 7.25×10⁻³ mol / 0.0015 kg = 4.83 m
- Freezing T° of solution = 1.86°C /m . 4.83 m - Freezing T° of pure solvent
-Freezing T° of solution = 1.86°C /m . 4.83 m - 0°C
Freezing T° of solution = - 8.98°C
