A solution of (solute) chloroacetic acid (C_2 H_3 O_2 Cl) is prepared in solvent

Question

A solution of (solute) chloroacetic acid (C_2 H_3 O_2 Cl) is prepared in solvent dichlorobenzene (C_6 H_4 Cl_2), where 0.942 g of chloroacetic acid is dissolved in 27.99 g of dichlorobenzene The freezing point depression constant (K_f) of the solvent dichlorobenzene is 7.271 degree C/m (where m represents molal) and the freezing point of the pure solvent is 53.00 degree C. The molar mass of chloroacetic acid is 94.50 g/mol. Given this information, what do you expect the freezing point depression (delta T_f) to be for this solution? Use this to figure out the expected freezing point (T_f) of the solution. expected delta T_f = _______ expected T_f = _____ Being a dedicated experimentalist, you decide to do verify these results in the lab, and find to your surprise that the freezing point is actually 51.68 degree C. Use this value to find the actual freezing point depression (delta T_f) for this solution. Actual delta T_f = ___________ You are confused, so you look up the value of the freezing point depression constant again, and find that indeed it is 7.271 degree C/m. Without understanding the results, you go ahead and calculate a number of particles per formula unit for chloroacetic acid (an "i" value). What is the calculated "i" value? From the value of i, can you conclude anything about the number of actual particles in solution? Looking at the structure of chloroacetic acid, can you justify this in any way? Make a sketch to show your point, and explain the calculated value of i (draw the sketch and type your explanation and on a separate page and attach it).

Explanation / Answer

1.

Freezing point depression = Kf*i*m

Where Kf= freezing point depression constant = 7.271 deg.c/m

And i= Van’t Hoff factor = 1 and m =molality= moles of solute/kg of solvent = (0.942/94.5)/27.99/1000=0.356

Freezing point depression = 7.27*0.356*1= 2.6 deg.c

Freezing point = 53-2.6= 51.4 deg.c

2. Freezing point actually the given case= 51.68

Freezing point depression = 53-51.68=1.32 deg.c

3. i= Freezing point depression/Kb*m= 2.6/(7.271*0.356)=1

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