Experiment 10 Colligative Properties ntroduction A salt will make boiling water
ID: 511110 • Letter: E
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Experiment 10 Colligative Properties ntroduction A salt will make boiling water hotter. A handful wil clear the snow from your steps, Every day we manipulate the physical properties of materials around us, but often don't know why use these particular chemicals or chemistry is involved. we 0boang adding a second chemical to the pure solvent are examples of manipulating colligative the boiling point of a In this lab, we wa study how varying the concentration and type of a satafects coligative properties are properties of a solution containing 2 or more components. Typically, the primary properties such as color or hardness, colligative properties d on the of solute in the solution, not the identity of the particles. In many cases, chemical properties of the the a colligative does not depend upon the nature of the particles. collgative commonly are point and freezing point depression (the temperature of freezing of a solution is lower than of a pure solventi. We wil be studying boang point elevation in this solute per liter of solution. point (as well as those for freezing point depression) use molalty, m, of solute per kilogram of solvent. By definton, boiling elevation involves a change of temperature. When the changes. Since molarity temperature of a solution changes, the volume of the solution also depends on the volume will Molality depends on the mass of the solvent, and thus is independent of temperature. dissociates, the in solution. the dissociates into and of the number of moles total number of particles the solution based on the of cations and anions produced from the dissociation. Thus Nack Na Charo (2moles of particles) Cacbno Ca w 2Chr amoles of particles) The quantity of particles produced per mole of solute is called the van't Hoff factor and is given the symbol. If 1 mole of sodium chloride were completely dissociated, i be 2. This theory works very wel for would dilute solutions and non-electrolytes (i 1, no dissociation at al since do not onize), but does not always apply to more concentrated solutions, Thus, when there are a large number of solution, they have a tendency as the salt and you dont get 100% dissociation For example, at a NaCl concentration of 0.1M iwil be 1.81 (90% dissociation) rather than 2 (00%). When a solution is heated, the presence of a non-volalle solute causes the solution to boil at a higher temperature than the pure solvent. The diference between the boiling point of the pure liquid and the boiling point of the solution is ATExplanation / Answer
Ans. #9. Given,
Molality of solution, m = 2.5 m
Volume of solution = 125 mL
#. 2.5 m means that there are 2.5 moles of NaCl per kg solvent.
So, moles of NaCl per kg solvent = 2.5 mol
Mass of NaCl per kg solvent = Moles of NaCl x Molar mass
= 2.5 mol x (58.442468 g/ mol)
= 146.11 g
Total mass of solution = Mass of solvent + Mass of solutes (NaCl)
= 1000 g + 146.11 g = 1146.11g
Volume of solution = Total mass / density
= 1146.11 g/ (1.02 g/ mL)
= 1123.63 mL
Therefore, 1123.63 mL of solution consists of 2.5 mol NaCl (= 146.11 g)
Now,
Required mass of NaCl for 125 mL solution =
(Mass of NaCl/ L solution) x Required volume of solution.
= (146.11 g / 1123.63 mL) x 125 mL
= 16.25 g
Ans. #10. Given,
0.5 m means that there are 0.5 moles of NaCl per kg (= 1000 g) solvent.
So, moles of NaCl per kg solvent = 0.5 mol
Mass of NaCl per kg solvent = Moles of NaCl x Molar mass
= 0.5 mol x (58.442468 g/ mol)
= 29.22 g
Total mass of solution = Mass of solvent + Mass of solutes (NaCl)
= 1000 g + 29.22 g = 1029.22 g
Volume of solution = Total mass / density
= 1029.22 g / (1.05 g/ mL)
= 980.21 mL = 0.98021 L
Therefore, 980.21 mL of solution consists of 0.5 mol NaCl.
Now,
Molarity of solution = Moles of NaCl / Volume of solution in liters
= 0.5 / 0.98021 L
= 0.510 mol/ L ; [1 mol/ L = 1 M]
= 0.510 M
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