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what are emulsions ? write two examples of emulsification.

Emulsion is the type of colliodal mixture :- In which the dispersion medium is oil and dispersed phase is water. for example ‘Butter’ In which the dispersion medium is water and dispersed phase is oil. for example ‘milk’ . Applications or examples of emulsification In case of cleaning action of soap, the soap solution acts

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A new beginning… 2022

We at wish you all have a very best time . All the sorrows , failures and bad moments be forgotten in the bright light of success and eternal love. May god bless you all. Best wishes The team Sarvpal Sharma, Suman Sharma, Ankush Sharma , Ramandeep Sharma, Karan Sharma , Shehanshah and Jaidev

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Give reason for why the enthalpy in case of chemisorption is usually higher than that of physisorption ?

In case of chemisorption there is a partial bond like interaction between the adsorbate and adsorbent . During such an interaction the higher amount of energy is required because bond formation and breaking will require more energy. but on the other hand in case of physisorption the adsorbate and adsorbent particles are simply held together

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Why is freezing point depression of 0.1 M sodium chloride solution is nearly twice that of 0.1 M glucose solution ?

the depression in freezing point (which is a colligative property) depends upon the number of solute particles present in the solution. when sodium chloride is mixed in the water it dissociates into Na+ and Cl– ions thus this increases the number of particles . but glucose don’t dissociate and its freezing point is higher than

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A 0.1539 molal aqueous solution of cane sugar (molar mass = 342g )  has a freezing point of 271K while the freezing point of pure water is 273.15K . what will be the freezing point of an aqueous solution containing 5g of glucose (molar mass= 180g) per 100 g of solution?

Molality of the given solution = 0.1539m ∆Tf = 273.15 – 271 = 2.15 K Therefore ∆Tf  = Kf .m Kf = ∆Tf / m Kf = 2.15/0.1539  Now mass of the solute W2 = 5g Molar mass of the solute M2 = 180g Mass of the solution W1 = 100g  Now the mass of

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

Solution:- Mass of ethylene glycol (solute) = 222.6 g Molar mass of ethylene glycol ( C2H4(OH)2 = 62g Number of moles of the solute = 222.6 / 62  = 3.59 moles Mass of the solvent (water) in solution = 200g = 0.200Kg Volume of the solution = 422.6 / 1.072 = 394.2 ml                                                 =

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Give an example each of miscible liquid pairs showing positive and negative deviations from Raoult’s law. Give reasons for each such deviation.

Solution The mixtures showing positive and negative deviations are as follows:- Mixtures showing positive deviation is mixture of ethanol and acetone. The reason behind this is :- in this case the solute and solvent interactions are weaker than the solute-solute and solvent-solvent interactions. Mixtures showing negative deviation is mixture of chloroform and acetone. The reason

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