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What is meant by positive and negative deviations from Raoult’s law and how is the sign of Δmix H related to positive and negative deviations Raoult’s law?

Answer:

According to Raoult's law, the partial vapour pressure of each component in any solution is directly proportional to its mole fraction.

The solution which obeys Raoult's law over enitre range  is known as ideal solution.
The solution which do not obeys Raoult's law is known as non- ideal solution.

Non-ideal solution have vapour pressure either higher or lower is  predicted by raoult's law

If the vapour pressure is higher then solution is said to exhibit positive deviation, And if the vapour pressure is lower than the solution than it said to be negtive deviation.

Positive Deviation from Raoult’s law. In those non-ideal solutions, when partial pressure of component ‘A’ in the mixture of ‘A’ and ‘B’ is found to be more than that calculated from Raoult’s law. Similarly, the partial vapour pressure of component ‘B’ can be higher than calculated from Raoult’s law.
This type of deviation from ideal behaviour is called positive deviation from Raoult’s law,
e.g., water and ethanol, chloroform and water, ethanol and CCl4, methanol and chloroform, benzene and methanol, acetic acid and toluene, acetone and ethanol, methanol and H2O.
For positive deviation ΔHmixing > 0. (+ ve)

Negative Deviation from Raoult’s law. When the partial vapour pressure of component ‘A’ is found to be less than calculated from Raoult’s law on adding the second component ‘B’. When A is added to B, the partial vapour pressure of solution is less than that of ideal solution of same composition. Boiling point of such a solution is relatively higher than the boiling point of A and B respectively. This type of deviation from ideal behaviour is known as negative deviation from Raoult’s law e.g., chlorofom and acetone, chloroform and methyl acetate, H2O and HCl, H2O and HNOacetic acid and pyridine, chloroform and benzene.
For negative deviation ΔHmixing < 0.

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A solution of glucose in water is labelled as 10% W/W, what would be the molality and mole fraction of each component in the solution? If the density of the solution is 1.2 g mL–1, then what shall be the molarity of the solution? 

Consider 100 g of 10% solution of glucose whose conc. is 10% W / W.
Mass of solution = 100 g
Mass of glucose = 10 g
Mass of solvent = 100 – 10 = 90 g
Molar mass of glucose,
C6H12O6 = 6 x 12 + 12 x 1 + 6 x 16
= 72 + 12 + 96 = 180 g / mol

Molality = Mass of soluteMolar mass of solute×1Mass of solvent in kg
            = 10180×0.09 kg = 0.617 m.

     Moles of H2O, nA = 9018 = 5
           Moles of glucose, nB = 10180 = 0.055
                      0.0555+0.055 = 0.0108 = .01
Mole fraction of glucose, xB = nBnA+nB
Mole fraction of water
            xA = 1-xb = 1-0.0108 = 0.989 = 0.99
                       Density of sol = 1.2 g mol L
Therefore, volume of solution
                 = MassDensity  = 1001.2 = 83.33 mL

                  Molarity  = 10/18083.33/100=0.67

 
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A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.

Mass of solute in 300 g of solution of 25% conc.
= 25100×300 = 75 g
Mass of solvent in 300 g of solution
= 300 – 75 = 225 g Moles of solute in 400 g of solution of 40% conc.
= 40100×400 = 160 g
Mass of solvent in 400 g of solution
= 400 – 160 = 240 g Total mass of the solute
= 75 + 160 = 235 g Total mass of solvent
= 225 g + 240 g = 465 g Total mass of solution
= 300 + 400 = 700 g Therefore, composition of solute in solution after mixing = 235700×100 = 33.57%
Percentage composition of solvent in solution after mixing
= 465700×100 = 66.43%

117 Views

State Henry’s law and mention some important applications.

Answer:

Henry’s law states that the solubility of a gas in a liquid is directly proportional to pressure of the gas; temperature constant.

Mathematically, P = KH x

where
x=mole fraction of gas in solution
P is partial pressure of gas,
KH is Henry's constant.

Applications:

(1) Henry's law finds various applications in industry and enables us to explain and understand some biological phenomena. The some of important applications are : CO2 solubility in soft drinks, beverages, soda water etc. is increased by applying high pressure and bottles are sealed under high pressure.

(2) For deep divers, oxygen diluted with less soluble He gas is used as breathing gas and it minimises the painful effects due to higher solubility of N2 gas in blood.

(3) In lungs, where oxygen is present in air with high partial pressure, haemoglobin combines with O2 to form oxyhaemoglobin. In tissues where partial pressure of O2 is low, oxyhaemoglobin releases the oxygen for utilisation in cellular activities.

 
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How many mL of 0.1 M HC1 are required to react completely with 1 g mixture of Na2COand NaHCO3 containing equimolar amounts of both?


Answer:

Let ‘a’ moles of Na2CO3 and a moles of NaHCO3 are present in 1 g equimolar mixture of two. Then we can write (a x 106) + (a x 84) = 1
(Molar masses are : Na2CO3 = 106 and NaHCO3 = 84)
a = 5.26 x 10–3 ‘a’ moles of Na2CO3 = 2
a equivalent of Na2CO3 ‘a’ moles of NaHCO3 
=a equivalent of NaHCO(2a x 1000) + (a x 1000) = 0.1 x V
∴                   3a = 0.1V1000 = 1×10-4V

[where V is volume of HCl (0.1 M)] Substituting the value of a, we have
1 x 10–4 = 3a = 3 x 5.26 x 10–3
V = 157.8 mL

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