Question

Explain your answer in terms of LeChatelier's principle. c) The above solution is divided equaly into three test tubes A, B. and Test tube "A is placed in an ice bath kept at 0 "C, test tube "B is kept at room temperature (25 °C) and test tube "C is placed in a boiling hot water bath maintained at 100 "C. The folowing are the colors of the solutions observed in each case. Solution "A" remained pink, solution "B" became dark pink and solution "C turned blue. What do you conclude about the effect of temperature change on the position of equilbrium between cobalt complex ions (use the equation given above). Is the forward equilibrium reaction an expthermic or endothermic process? Why? Explain your answer cleanY 5. The following table represents the solubility (g/100 mL H2O) of salt X at different temperatures. 60.0 80.0 102.0 20.0 40.0 Temperature (C) 32.0 64.0 94.0 130.0 46.0 Solubility (p/100 mL H,O) a) Plot the solubility curve for salt X in the grid provided. (Properly label the graph) Using the solubility- temperature plot to answer the following questions: b) Determine the solubility of the salt X in water from the drawn graph at 75.0 c2 c) if you weigh out 25.0 g of salt X into 50.0 mL. of water, what will be its solubility? d) Find the saturation temperature required to dissolve all the solid completely mentioned in (c)

Answer 1

Anss 01 (ou to apply hnignos 50 nnt DSL 10 Temperature coc)

Expt. No Page No. 1 Date Given data Temperature (°C) solubility (g/100mL ho) 20.0 32.0 40.0 60.0 46.0 64.0 80.0 94.0 102.0 130.0 from given data and graph it is clear that solubility of salt x incleases with increase in temperature It means salt dissolves in water endothermally This can be explained by Le-Chatelier's principle - Xcs) tho Xeag) An= the According to Le-chatelier's principle with increase in temperature, the reaction proceeds. in a direction where then this excess heat can be absorbed to maintain the equillibrium (6 from graph At 75c solubility of -=839/ 100 ml ho salt x dissolved (c) Soubility is expressed as gram of solute in loog of solvent. Density of water I glums 1. mass of water z volume of water Teacher's Signature:...

Giren : Mass of salt x = 25.0g. Volume of water = 5000 ml Solubility of salt x = Giren mass of solute - 100 (g/100 mL ho) Volume of water Il 25.0 (9) x 500 so o (me) = 50 g / 100 m2 to - sog / 100 g ho id saturated solutions are the solutions in which no more solute can be dissolved. OR saturated solutions are solutions having maximum of amount of solute an dissolved form for part (1) solubility of salt x = Sog / 100ml to

Expt. No. Page No. 3 Date from graph, temperature corresponds to solubility sog/100 mlho is 44°c. Thus saturation temperature for part (1) = 44°C Above this unsaturated temperature, the solution becomes salt I for example dissolved in At 60°, mass of 100 mL of yo = 64 g that can be (from graph) for past (c) mass of salt = sog Difference in masses a 64-50 = 149, it means after dissolving so g in 100 mL of ho, another of salt can be dissolved without precipitation A Below this temperature (44°c), the excess of salt get precipitates out. for example at 40°C, mass of salt that can be dissolved in 100 mL of ko = 46g (from graph) Teacher's Signature :

Page 4 ence in masces - for part (c) mass of salt = sog. Difference in masses = 50 - 46 = 4g. It means, at 40°C 46g of salt dissolres completely in 100 mL of to, but this extra 4g will precipitate out.