Question

Match the following aqueous solutions with the appropriate letter from the column on the right. 1. 0.17 m Mg(CH3COO)2 A. Lowest freezing point 2. 0.11 m Cr(CH3COO)3 B. Second lowest freezing point 3. 0.24 m KBr C. Third lowest freezing point 4. 0.53 m Sucrose(nonelectrolyte) D. Highest freezing point Submit Answer Retry Entire Group

Answer 1

Step-by-step explanation:

• Using the Vant Hoff equation

• Mathematically ; ΔTf = i x kf x m

• where ΔTf = depression in freezing point

• i = vant hoff factor

• kf = freezing point depression constant

• m = molality

The lesser the value of ΔTf , the higher the freezing point

for 0.17 m Mg(CH3COO)2, i =2, ΔTf = i x m = 2 x 0.17 = 0.34

for 0.11 m Cr(CH3COO)3, i = 3, ΔTf = 3 x 0.11 = 0.33

for 0.24 m KBr, i = 1, ΔTf = 1 x 0.24 = 0.24

for 0.53 m Sucrose(non-electrolyte) , i = 1, ΔTf = 1 x 0.53 = 0.53

From the values of the of the depression in freezing point gotten ;

KBr - Highest freezing point

Cr(CH3COO)3 - Second lowest freezing point

Mg(CH3COO)2 - Third lowest freezing point

Sucrose(non-electrolyte) - Lowest freezing point