Which of the following solutions will have the lowest freezing point? a) 0.015 m mgcl2 b) 0.0100 m nacl c) 0.035 m ch3ch2ch2oh d) 0.0100 m li2so4

A) 0.015 m MgCl₂.

Explanation:

ΔTf = i.Kf.m,

where, ΔTf is the depression in freezing point.

i is the van 't Hoff factor.

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.

Kf is the molal depression constant of water (Kf = 1.86°C/m).

m is the molality of the solution.

A) 0.015 m MgCl₂:

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.

B) 0.01 m NaCl:

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.

C) 0.035 m CH₃CH₂CH₂OH:

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.

D) 0.01 m Li₂SO₄:

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.

po-215 i got it right on the quiz