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Lab 3: Determination of Molar Mass by Freezing Point Depression 1 During winter, several different "salts"...


Lab 3: Determination of Molar Mass by Freezing Point Depression 1 During winter, several different salts are used to melt i
Lab 3: Determination of Molar Mass by Freezing Point Depression 1 During winter, several different "salts" are used to melt ice sodium chloride (NaCI), a mix of sodium chloride and magnesium chloride (NaCl and MgCla calcium magnesium acetate (CaMg(CHO)). Assume you have 1 mole of each of these in 1 kg of water. Compare the freezing point depressions of these three "salts." a. b. What is the ratio of moles of "salt" to freezing point depression for each? What is the ratio of grams of "salt" to freezing point depression for each? 01 C. Error analysis: If some of the solute was lost as you transferred it, what impact would that have on oe 2. (Hint compare the result you would get to the result without this error) the moles of solute? Explain. a the molality? Explain. b. the freezing point of the solution? Explain. C d. the molar mass? Explain.
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a)

Freezing point depression constant for the the water is 1.86 K.Kg/mole

molality of the the solution is

For NaCl is 1 Mole/Kg of solution

For Mixture of NaCl and MgCl2 is 2 Mole/Kg of solution

For the calcium magnessium acetate is 1 Mole/Kg of solution

the molality is same in each of the three cases so the depression in the freezing point is same for all cases

AT = K* m

Kf is the freezing point depression constant

m is the molality of the solution

freezing point of pure water is 273.15 K

К.Кg AT = 1.86A.Kg Mole 1 Mole Ко

hence the depression in the freezing point is 1.86 K for the NaCl and Calcium Magnesium Acetate solution

new freezing point of the solution is 273.15-1.86=271.29 K

For MgCl2 and NaCl both are one mole  

= 1.86*2 = 3.72 K

the new freezing point is 273.15 - 3.72 = 269.43 K

for the case where all three are present in the solution hence the concentration of the solution now becomes 3 moles/Kg

new freezing point depression is

=1.86*4 K

= 7.44 K

new freezing point is 273.15 - 7.44 = 265.71 K

b)

ratio of moles of salts to the boiling point depression

1.) For NaC 1.86

2.) For NaClandMg Cl2 3.72

3.) ForCalciumMagnessium Acetate 1.86

4 4.) for AlThree 7.44

c)

58.5 1.) For Nacl 1.86

153.5 2.) For NaClandMgCl2= 3.72

300 3.)ForCalciumMagnesiumAcetate 1,86

453.5 4.) For AllThree 7.44

Answere no 2

a)

As the only solute is lost hence only solute moles will decrease

b)

As the solvent is not lost only solute is lost hence molarity as well as molality will decrease

c)

As the molality is decreasing because of decrease in moles of the solute so freezing point depression will also decrease

d)

molar mass is the constant quantity so the it will not decrease because molar mass for one mole of substance is constant  

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