Lyophilic sols are more stable than lyophobic sols. Why ?

Lyophilic sols are more stable than lyophobic sols.this is due to the fact that lyophilic colloids are extensively solvated, i.e. colloidal particles are covered by a sheath of the liquid in which they are dispersed.

Lyophilic colloids have a unique property of protecting lyophobic colloids. When a lyophilic sol is added to the lyophobic sol, the lyophilic particles form a layer around lyophobic particles and, thus, protect the latter from electrolytes.

Lyophilic colloids used for this purpose are called protective colloids.

Give some examples of common negatively charged colloidal solution?

The common negatively charged colloids are given below:

As2S3 sol,

Sb2S3 sol, 

CdS sol, 

Au sol, 

Cu sol, 

Ag sol 

And acid dyes like

congo red.

Give some examples of common positively charged colloidal solution?

The common positively charged colloids are given below:

Fe(OH)3 sol, 

Cr(OH)3 sol,

Al(OH)3 sol, 

Ca(OH)2 sol, 

And also dyes like methylene blue and haemoglobin.

A mixture of weighing 228 g contain CaCl2 and NaCl . If this mixture is dissolved in 10 kg of water and form ideal solution that boil at 100.364 ℃ The mol % of NaCl in mixture is [ kb of water = 0.52 K per mol Kg]

Given that 228 g mixture of NaCl and CaCl₂ consider that “x” g NaCl and (228-x) gm CaCl₂ present in this mixture:

NaCl and CaCl₂ both are strong electrolytes hence Vant’s Hoff factor are 2 and 3 respectively.

Moles of NaCl = x/58.5

Moles of CaCl₂ = 228-x

And hence molality of NaCl and CaCl₂ are 

Molality of NaCl = ×/58.5×10kg

Molality of CaCl2= 228-x/111×10kg

We know that 

∆Tb = i₁m₁+i₂m₂

100.364= 2(×/58.5×10kg) +3(228-x/111×10kg)

On solving we got x= 117 g 

Moles of NaCl= 117/58.5 =2

Moles of CaCl2= 111/111=1

Mole percentage (%) 0f NaCl = (2/2+1)x100= 66.67%

Related Questions:

(1) An aqueous solution has 5% urea and 10% glucose by weight. what will be the freezing point of this solution ? ( Kf = 1.86 K Kg per mole)

(2) Acetic acid dimerised to an extent of 30% in benzene . The observed molecular mass (in gm) is...

(3) How much ice will separate if a solution containing 25 g of ethylene glycol [C2H4(OH) 2] in 100g of water is cooled to 10°C? Kf(H2O) = 1.86 (25.05 g)

(4) The van't Hoff's factor for 0.1 M Ba(NO3)2 solution is 2.74. the degree of dissociation is.

(5) Aqueous solution of 0.004 M Na2SO4 and 0.01 M glucose are iosotonic .The degree of dissociation of Na2SO4 is ..

(6) A saturared solution of XCl3 has a vapour pressure 17.20 mm Hg at 20°C, while pure water vapour pressure is 17.25 mm Hg. Solubility product (Ksp) of XCl3 at 20°C is

What is fractional presence of Fe+2 in Fe0.94 O1.0 ?

Fe0.94 O 1.0 is non stachiometric compound hence Nickel contains both Fe+2 and Fe+3 .

We can calculate ℅  presence of Fe+2  and Fe+3 by oxidation method.

(F+2+Fe+3) O1.O

2x+3(0.94-x)-2=0

Fe+2=x=0.82 or 82% and Fe+3= 8%