Predicting relative boiling point elevations and freezing point dep. Four liquid
ID: 717051 • Letter: P
Question
Predicting relative boiling point elevations and freezing point dep. Four liquids are described in the table below. Use the second column of the table to explain the order of their freezing points, and the third column to explain the order of their boiling points For example, select '1' in the second column next to the liquid with the lowest freezing point. Select '2' in the second column next to the liquid with the next higher freezing point, and so on. In the third column, select '1 next to the liquid with the lowest boiling point, 2' next to the liquid with the next higher boiling point, and so on. Note: the density of water is 1.00 g/mt. solution freezing point boiling point 6.1 g of potassium hydroxide (KOH) dissolved in 150. mL of water se one) 6.1 g of ethylene glycol (CHg02) dissolved in 150. mL of water 6.1 g of sucrose (C1zHz201) dissolved in 150. mL of water 150. mL of pure water (choose one). (choose one)(choose one) choose one) . (choose one) (choose one) (choose one)Explanation / Answer
Answer:
Pure water will have a boiling point of 100 deg.c and freezing point of 0 deg.c.
for water Kb= 0.512 deg.c/m and kf= 1.86c/m
Boiling point elevation = kb*m*i,
where i is Vant Hoff factor, m = molality = moles of solute/ kg of solvent.
water density is assumed as 1 g/cc,
so, mass of water = 150 ml* density =150 gm= 0.15 Kg.
Moles = mass/Molecular weight
molecular weight of KOH = 56,
Molecular weight of C2H5O2 = 62 and
Molecular weight of C22H12O11 = 342.1
noles of KOH = 6.1 / 56
moles oh ethylene glycol = 6.1 / 62
moles of sucrose = 6.1 / 342
now,
molarity of KOH = (6.1 / 56) / .15 L
molarity of ethylene glycol = (6.1 / 62) / .15 L
molarity og sucrose = (6.1 / 342) / .15 L
-----------------------------------------------------------------
now,
boiling point elevation = Kb *m*i
so,
boiling point elevation of KOH = 0.512 * (6.1 / 56) / .15 L = 0.371
boiling point elevation of ethylene glycol = 0.512 *(6.1 / 62) / .15 L = 0.335
boiling point elevation of sucrose = 0.512 * (6.1 / 342) / 9.15 L = 0.0608
--------------------------
boiling point of (in degree)
solution of KOH = 100 + 0.371 = 100.371
solution of erhylene glycol = 100 + 0.335 = 100.335
solution of sucrose = 100 + 0.0608 = 100.0608
so , order of boiling point is water, KOH, ethylene glycol, sucrose
-----------------------------------------------------------------------------------------------
freezing point depression = Kf * m * i
freezing point depression of
KOH = 1.86*((6.1 / 56) / .15 L)*1 = 1.35
ethylene glycol = 1.86*((6.1 / 62) / .15 L )* 1 = 1.22
sucrose = 1.86 * ((6.1 / 56) / .15 L )* 1 = 0.22
----------------------
boiling point of (in degree)
KOH solution = 0 - 1.35 = - 1.35
ethylene glycol = 0 - 1.22 = - 1.22
sucrose = 0 - 0.22 = - 0.22
orddr of freezing point is
sucrose, ethylene gylcol, KOH , water
-