Complete the following table One may now prove, as was done for the ring Z, that
ID: 890443 • Letter: C
Question
Complete the following table
One may now prove, as was done for the ring Z, that L(a(x).b(x)) = L(g(x).0) = {m(x)g(x) |m(x) F[x]}, where g(x) made monic is by definition the greatest common divisor (a(x),b(x)). Instead construct the greatest common divisor as follows. Let h(x) denote a non-zero polynomial of lowest degree in L(a(x),b(x)) (it is at this point possible there are many such, even if F has only two elements). Then first prove, using only the definition of L(a(x),b(x)), not 2., that L(a(x),b(x)) = {m(x)h(x) | m(x) F[x]}. (.Suggestion: Let f(x) L(a(x),b(x)). By DA there are q(x) and r(x) such that f(x) = h(x)q(x) + r(x), where deg r(x)Explanation / Answer
Sodium sulfate:
Moles of Sodium sulfate = Mass of solute / Molar mass = 12 g / 142 g mol-1 = 0.0845 mol
Mass of water = Moles of solute / Molality = 0.0845 mol / 0.12 mol kg-1 = 0.704 kg = 704 g
Expected freezing point = Freezing point of water - [Molality x molal freezing point constant of water]
= 0 C - [0.12 mol kg-1 x 1.86 C kg mol-1] = 0 - 0.22 C = - 0.22 C
Magnisium nitrate:
Molality = [Freezing point of water - Expected freezing point] / molal freezing point constant of water
= [ 0 C - (-3.0 C)] / 1.86 C kg mol-1 = 1.613 mol kg-1
Mass of solute = Mass of water x Molality = 0.25 kg x 1.613 mol kg-1 = 0.403 mol x 148.3 g mol-1 = 59.76 g
Glucose:
Mass of solute = Mass of water x Molality = 0.12 kg x 0.28 mol kg-1 = 0.0336 mol x 180 g mol-1 = 6.048 g
Expected freezing point = Freezing point of water - [Molality x molal freezing point constant of water]
= 0 C - [0.28 mol kg-1 x 1.86 C kg mol-1] = 0 - 0.52 C = - 0.52 C
Unknown:
Molality = [Freezing point of water - Expected freezing point] / molal freezing point constant of water
= [ 0 C - (-1.62 C)] / 1.86 C kg mol-1 = 0.87 mol kg-1
Moles of solute (unknown) = Mass of water x Molality = 0.075 kg x 0.87 mol kg-1 = 0.065 mol
Molar mass = Mass of solute / Moles of solute = 5.3 g / 0.065 mol = 81.54 g mol-1