Question

just 2-9 please

Pre-Lab Questions: Freezing Point Depression Name Lab Section Read the description of the experiment. Make sure you understand it before you go any further 1. Graph the following data (use graph paper and a ruler). You can get graph paper from the lab, or print it from a web site. Draw best-fitnes really best fit curves). They will be similar in shape to these shown on the graph in the "Background" section. Use different colors to draw data points and curves for water and NaCl solution. Apply explanation provided for the figure in "Background" section to determine the freezing point depression of the NaCl solution Water NaCl solution Time (s) Temp °C Time (s) Temp °C 13.4 13.2 4.6 -4.1 -0.7 120 150 180 0.4 120 150 180 0.5 210 0.3 240 210 240 0.4 270 300 0.4 0.2 0.4 270 300 330 360 330 360 -0.1 -0.1 Freezing point depression (At): - Freezing point NaCl solution Freezing point Water: 2. What is the freezing point depression (AT) of water in a solution of 17.1 g of sucrose, C2H2O., in 200.0 g of water? What is the actual freezing point of the solution in C? Calculations: Freezing point of the solution = _ _°C Freezing point depression (AT) = next page 3. What is the freezing point of 1 m acetic acid (note: it is not pure acetic acid, but a water solution of the acetic acid). What, if any, assumptions do you need to make?

Answer 1

point LoTf) is a AL-2 Deppresion in freezing 1. colligative property; It is calculated by (oTf) руу = kxm r where m=mo e m=molality - w buzion] - =ky hsolute J m= hsolute Mseluent inkg M selvent iniy = 1.86 X 17.1 hsalute=W 399 goo Tono = 17.1 392 of) Moleular weight of o Te = 1.86 X 17.1X1000 - 3924200 lotus = 0.405°C 1 sucrose; 42 Maa = 12412 +22*1 +1876 Depression in freezing point=0.405°C/ 9.900 = 144+22+176 Actual freezing point of 1 = 392 the 14 = 1.86 # met solution) ot; = T10-T 17 och med Z V stardazol freezing point of the solution; (To=0) 0.405 =0 T(T) = -0.405c7 ferezing point of the solution, -0.405°C] Actual