sucrose like many sugar is highly soluble in water almost 2000 g will dissolve in1 litre of water giving rise to what amount pancake syrup estimate the boiling point of sugar solution?
Answers
Answer:
ANSWER
DepressioninFreezingpointΔT
f
=
M
1
×W
2
W
1
×K
f
×1000
whereW
1
=Weight of Solute
W
2
=Weight of solvent
M
1
=Molar mass of solute
K
f
=Freezing point deprssion constant
Now,ΔT
f
=
342×1000
1.86×68.5×1000
=0.372C
Now,ΔT
f
=T
o
−T
f
So,T
f
=0−0.372=−0.372C.
(Freezing point of purewater=0
0
C.)
Answer:
The colligative properties that we will consider in this and the next unit apply to to solutions in which the solute is non-volatile; that is, it does not make a significant contribution to the overall vapor pressure of the solution. Solutions of salt or sugar in water fulfill this condition exactly. Other solutes that have very small vapor pressures, such as iodine or ethylene glycol antifreeze, can often be considered nonvolatile in comparison to the solvent at the same temperature. Solutions in which both components possess significant vapor pressures, such as alcohol in water, will be treated in another section farther on.
1 Vapor pressure of solutions: Raoult's law
The colligative properties really depend on the escaping tendency of solvent molecules from the liquid phase. You will recall that the vapor pressure is a direct measure of escaping tendency, so we can use these terms more or less interchangeably.
The tendency of molecules to escape from a liquid phase into the gas phase depends in part on how much of an increase in entropy can be achieved in doing so. Evaporation of solvent molecules from the liquid always leads to a large increase in entropy because of the greater volume occupied by the molecules in the gaseous state. But if the liquid solvent is initially “diluted“ with solute, its entropy is already larger to start with, so the amount by which it can increase on entering the gas phase will be less. There will accordingly be less tendency for the solvent molecules to enter the gas phase, and so the vapor pressure of the solution diminishes as the concentration of solute increases and that of solvent decreases.
Raoult's law
The number 55.5 mol L–1 (= 1000 g L–1 ÷ 18 g mol–1) is a useful one to remember if you are dealing a lot with aqueous solutions; this represents the concentration of water in pure water. (Strictly speaking, this is the molal concentration of H2O; it is only the molar concentration at temperatures around 4° C, where the density of water is closest to 1.000 g cm–1.)
Diagram 1 (above left) represents pure water whose concentration in the liquid is 55.5 M. A tiny fraction of the H2O molecules will escape into the vapor space, and if the top of the container is closed, the pressure of water vapor builds up until equilibrium is achieved. Once this happens, water molecules continue to pass between the liquid and vapor in both directions, but at equal rates, so the partial pressure of H2O in the vapor remains constant at a value known as the vapor pressure of water at the particular temperature.
In the system on the right, we have replaced a fraction of the water molecules with a substance that has zero or negligible vapor pressure — a nonvolatile solute such as salt or sugar. This has the effect of diluting the water, reducing its escaping tendency and thus its vapor pressure.
Raoult's law plotWhat's important to remember is that the reduction in the vapor pressure of a solution of this kind is directly proportional to the fraction of the [volatile] solute molecules in the liquid — that is, to the mole fraction of the solvent. The reduced vapor pressure is given by Raoult's law (1886):
From the definition of mole fraction (denoted by X), you should understand that in a two-component solution (i.e., a solvent and a single solute),
Xsolvent = 1–Xsolute
Problem Example 1
Estimate the vapor pressure of a 40 percent (W/W) solution of ordinary cane sugar (C22O11H22, 342 g mol–1) in water. The vapor pressure of pure water at this particular temperature is 26.0 torr.
Solution: 100 g of solution contains (40 g) ÷ (342 g mol–1) = 0.12 mol of sugar and (60 g) ÷ (18 g mol–1) = 3.3 mol of water. The mole fraction of water in the solution is
and its vapor pressure will be 0.96 × 26.0 torr = 25.1 torr.
Since the sum of all mole fractions in a mixture must be unity, it follows that the more moles of solute, the smaller will be the mole fraction of the solvent. Also, if the solute is a salt that dissociates into ions, then the proportion of solvent molecules will be even smaller.
Problem Example 2
The vapor pressure of water at 10° C is 9.2 torr. Estimate the vapor pressure at this temperature of a solution prepared by dissolving 1 mole of CaCl2 in 1 L of water.
Solution: Each mole of CaCl2 dissociates into one mole of Ca2+ and two moles of Cl1–, giving a total of three moles of solute particles. The mole fraction of water in the solution will be
The vapor pressure will be 0.95 × 9.2 torr = 8.7 torr.
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