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An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H4(OH)2) and 200 g of water. Calculate the molality of the solution. If the density of this solution be 1.072 g ml–1, what will be the molarity of the solution?


Answer:

Molecular mass of glycol C2H4(OH)2 =62
Number of moles of C2H4(OH)2
= mass of glycol / molar mass of glycol

= 222.6/62=3.59  moles

Molality of solution =
number of moles of solute / mass of solvent in Kg  

Mass of solvent = 200g = 200/1000 Kg = 0.2 kg

Plug the values we get

Molality = 3.59/0.2

Molality = 17.95

Formula of molarity of solution = number of moles of solute / volume of solution in Kg
Formula of volume = mass / density
Volume = 422.6g/(1.072 g/ml)
Volume = 394.21 ml
Convert in liter   
Volume in liter = 394.21 ml /1000 liter  =0.394 liter 

Molarity =9.1M
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Show graphically how the vapour pressure of a solvent and a solution in it of a non-volatile solute change with temperature. Show on this graph the boiling points of the solvent and the solution. Which is higher and why?

Answer:

Elevation of boiling point with addition of non-volatile solute vapour pressure decrese and hence boiling point increase.

Let Tb0 be the boiling point of pure solvent and
Tb be the boiling point of solution. The increase in
the boiling point  Tb = Tb - Tb0  is known as
elevation of boiling point.

for dilute
solutions the elevation of boiling point (ΔTb) is
directly proportional to the molal concentration of
the solute in a solution. Thus

     ΔTb ∝ m 
or
   ΔTb = Kb m

Here m (molality) is the number of moles of solute dissolved in 1 kg of solvent and the constant of proportionality, Kb is called Boiling Point Elevation Constant or Molal Elevation Constant. The unit of Kb is K kg mol-1.If w2 gram of solute of molar mass Mis dissolved in w1 gram of solvent, then molality, m of the solution is given by the expression:

m = 1000 × w2M2 ×w1putting the value of molality in Tb = Kbmwe get Tb = kb × 1000×w2M2×w1


Thus, in order to determine M2, molar mass of the solute, known mass of solute in a known mass of the solvent is taken and ΔTb is
determined experimentally for a known solvent whose Kb value is known.

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What is osmotic pressure? How would you determine the molecular mass of solute with the help of osmotic pressure?

Answer:
osmotic pressure is proportional to the molarity, C of the solution at a given temperature T. Thus:
Π = C R T 
Here Π is the osmotic pressure and R is the
gas constant.
Π = (n2 /V) R T
Here V is volume of a solution in litres containing n2 moles of solute.
If w2 grams of solute, of molar mass, M2 is present in the solution, then
n2 = w2 / M2 and we can write,

πV= w2RTM2

Thus, knowing the quantities w2, T, Π and V we can calculate the molar mass of the solute.

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Derive an expression relating the elevation of boiling point to the amount of solute present in the solution.

 Answer:

the vapour pressure of a liquid increases with increase of temperature. It boils at the temperature at which its vapour pressure is equal to the atmospheric pressure. For example, water boils at 373.15 K (100° C) because at this temperature the vapour pressure of water is 1.013 bar (1 atmosphere)

Let Tb0 be the boiling point of pure solvent and
Tb be the boiling point of solution. The increase in
the boiling point Tb = Tb - Tb0  is known as
elevation of boiling point.

for dilute solutions the elevation of boiling point (ΔTb) is directly proportional to the molal concentration of the solute in a solution. Thus
ΔTb ∝ m 
or ΔTb = Kb m
Here m (molality) is the number of moles of solute dissolved in 1 kg of solvent and the constant of proportionality, Kb is called Boiling Point Elevation Constant or Molal Elevation Constant (EbullioscopicConstant). The unit of Kb is K kg mol-1. If w2 gram of solute of molar mass M2 is dissolved in w1 gram of solvent, then molality, m of the solution is given by the expression:

m = w2M2w11000  = 1000×w2M2×w1Substituting the value of molality in Tbm we get Tb = Kb×1000×w2M2×w1M2 =  Kb×1000×w2Tb×w1

Thus, in order to determine M2, molar mass of the solute, known mass of solute in a known mass of the solvent is taken and ΔTb is
determined experimentally for a known solvent whose Kb value is known.

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Explain with a suitable diagram and appropriate examples. Why some non-ideal solutions show positive deviation from ideal behavior?

In case of positive deviation from Raoult's law, A-B interactions are weaker than those between A-A or B-B, i.e., in this case the intermolecular attractive forces between the solute-solvent molecules are weaker than those between the solute-solute and solvent-solvent molecules. This means that in such solutions, molecules of A (or B) will find it easier to escape than in pure state. This will increase the vapour pressure and result in positive deviation. Mixtures of ethanol and acetone is good example of this.

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