Answer:
A) 0.015 m MgCl₂.
Explanation:
- Adding solute to water causes depression of the boiling point.
- The depression in freezing point (ΔTf) can be calculated using the relation:
<em>ΔTf = i.Kf.m,</em>
where, ΔTf is the depression in freezing point.
i is the van 't Hoff factor.
<em>van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.</em>
Kf is the molal depression constant of water (Kf = 1.86°C/m).
m is the molality of the solution.
<u><em>A) 0.015 m MgCl₂:</em></u>
i for MgCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.
m = 0.015 m.
Kf = 1.86°C/m.
∴ ΔTb for (Li₂SO₄) = i.Kf.m = (3)(1.86°C/m)(0.015 m) = 0.0837°C.
∴ The freezing point of 0.015 m MgCl₂ = 0.0°C - 0.0837°C = - 0.0837°C.
<u><em>B) 0.01 m NaCl:</em></u>
i for NaCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.
m = 0.01 m.
Kf = 1.86°C/m.
∴ ΔTb for (NaCl) = i.Kf.m = (2)(1.86°C/m)(0.01 m) = 0.0372°C.
∴ The freezing point of 0.01 m NaCl = 0.0°C - 0.0372°C = - 0.0372°C.
<u><em>C) 0.035 m CH₃CH₂CH₂OH:</em></u>
i for CH₃CH₂CH₂OH = no. of particles produced when the substance is dissolved/no. of original particle = 1/1 = 1.
m = 0.035 m.
Kf = 1.86°C/m.
∴ ΔTb for (CH₃CH₂CH₂OH) = i.Kf.m = (1)(1.86°C/m)(0.035 m) = 0.0651°C.
∴ The freezing point of 0.035 m CH₃CH₂CH₂OH = 0.0°C - 0.0651°C = - 0.0651°C.
<u><em>D) 0.01 m Li₂SO₄:</em></u>
i for Li₂SO₄ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.
m = 0.01 m.
Kf = 1.86°C/m.
∴ ΔTb for (Li₂SO₄) = i.Kf.m = (3)(1.86°C/m)(0.01 m) = 0.0558°C.
∴ The freezing point of 0.01 m Li₂SO₄ = 0.0°C - 0.0558°C = - 0.0558°C.
So, the solution that has the lowest freezing point is: A) 0.015 m MgCl₂.
