engage Learning molar mass potassium bromide-Googl [Review Topic Referencen Use

Question

engage Learning molar mass potassium bromide-Googl [Review Topic Referencen Use the References to access important values if needed for this question. Match the following aqueous solutions with the appropriate letter from the column on the right. 1. 0.21 m NH4CI 2. 0.24 m NaCH COO 3. 0.15 m MnCl2 4. 0.51 m Sucrose(nonelectrolyte) A. Lowest freezing point B. Second lowest freezing point C. Third lowest freezing point D. Highest freezing point Submit Answer Retry Entire Group 9 more group attempts remaining

Explanation / Answer

Ans :

The freezing point depression = i. kf. m

here i = vanthoff's factor and m is molality , which determine the freezing point of the solutions.

So here :

1) 0.21 m NH4Cl = 2 x 0.21 x kf = 0.42 kf = highest freezing point

2) 0.24m NaCH3COO = 2 x 0.24 x kf= 0.48 kf = second lowest freezing point

3) 0.15 m MnCl2 = 3 x 0.15 x kf = 0.45 Kf = third lowest freezing point

4) 0.51m sucrose = 1 x 0.51 x kf = 0.51 Kf = lowest freezing point

Likewise :

1) 0.14 m Mn(NO3)2 = 3 x 0.14 x kf = 0.42 Kf = highest freezing point

2) 0.17 m Pb(NO3)2 = 3 x 0.17 x kf = 0.51 kf = lowest freezing point

3) 0.16 m (NH4)2SO4 = 3 x 0.16 = 0.48 kf = second lowest freezing point

4) 0.45 m sucrose = 1 x 0.45 x kf = 0.45 Kf = third lowest freezing point

here :

1) 0.16 m Ca(CH3COO)2 = 3 x 0.16 x kf = 0.48 kf = second lowest freezing point

2) 0.17 m Cu(CH3COO)2 = 3 x 0.17 x kf = 0.51 kf = lowest freezing point

3) 0.11 m AlBr3 = 4 x 0.11 x kf = 0.44 Kf = third lowest freezing point

4) 0.42 m sucrose = 1 x 0.42 x kf = 0.42 kf = highest freezing point

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