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When dissolved in 1000 g of water, which chemical compound will produce a solution with the greatest freezing point depression?

A5.0 g of sodium chloride (NaCl)
B9.2 g of potassium chloride (KCl)
C11 g of calcium fluoride (CaF2)
D20 g of glucose (C6H12O6)

Verified Answer :

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The correct answer among the choice presented above would be option C. The solution that will have the greatest freezing point depression is 11 g of calcium fluoride (CaF2). This is because it has higher concentration among other solutions. Also, when it ionizes 3 ions are formed.

Answer:

c) 11 g of calcium fluoride (CaF2)

Explanation:

Given :

Mass of solvent = M = 1000 g = 1 kg

Formula:

The depression in freezing point is given as:

\(\Delta T = k_{f}*m\)

where kf = freezing point depression constant for solvent

\(m = molality = \frac{moles of solute}{kg solvent}\)

\(moles = \frac{mass}{molar mass}\)

Calculation:

a) Mass of NaCl = 5.0 g

Molar mass = 58.44 g/mol

\(Moles NaCl = \frac{5}{58.44} =0.0856\)

\(Molality(NaCl) = \frac{0.0856 moles}{ 1 kg water} = 0.0856 m\)

\(\Delta T = k_{f}*0.0856\)

b) Mass of KCl = 9.2 g

Molar mass = 74.55 g/mol

\(Moles KCl = \frac{9.2}{74.55} =0.1234\)

\(Molality(KCl) = \frac{0.1234 moles}{ 1 kg water} = 0.1234 m\)

\(\Delta T = k_{f}*0.1234\)

c) Mass of CaF2 = 11.0 g

Molar mass = 78 g/mol

\(Moles CaF2= \frac{11}{78} =0.1410\)

\(Molality(CaF2) = \frac{0.1410 moles}{ 1 kg water} = 0.1410 m\)

\(\Delta T = k_{f}*0.1410\)

d) Mass of glucose = 20.0 g

Molar mass = 180 g/mol

\(Moles glucose = \frac{20}{180} =0.1111\)

\(Molality(glucose) = \frac{0.1111 moles}{ 1 kg water} = 0.1111 m\)

\(\Delta T = k_{f}*0.1111\)

Therefore, CaF2 will produce the greatest freezing point depression

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