Ncert Solutions Chemistry Class 12th

According to Henry's law, "At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid."
Stated as,
p = K_Hx 
Where,  
P = partial pressure of the solute above the solution
K_H = Henry's constant
x = concentration of the solute in the solution
Given,

K_H = 1.67x10⁸ Pa 

P_CO2  2.5 atm = 2.5 * 1.01325 * 10⁵Pa

According to Henry's law,

p = K_Hx

x = P/K_H

x = 2.5 X 1.01325 X 10⁵ / 1.67 X 10⁸

x = 0.00152

In 500 ml of soda water there is 500 ml of water (neglecting soda)

Mole of water = 500/18

= 27.78 mol

Now,

x = n_co2 / n_co2 + n_H2o

n_co2  = x n_H2o

 = 0.00152 X 27.78

= 0.042 mol

Hence, quantity of CO₂ in 500mL of soda water  0.042 * 44 = 1.848 g

According to Henry's law,"At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid."
Stated as,  
p = K_Hx 
Where, P = partial pressure of the solute above the solution
K_H = Henry's constant
x = concentration of the solute in the solution
Given,
Solubility of H₂S in water at STP is 0.195 m
We know,
At STP pressure p = 0.987 bar
0.195 mol of H₂S is dissolved in 1000g of water

Moles of water = 1000/18
 = 55.56 g/mol
∴ the mole fraction of H₂S = Moles of H₂S / Moles of H₂S + Moles of water
 = 0.195 / 0.195+55.6
 = 0.0035
According to Henry's law, p = K_Hx 
K_H = p/x
K_H = 0.987 / 0.0035
K_H = 282 bar 
∴ The Henry's law constant is 282 bar

(a) Molality, also called molal concentration, is a measure of the concentration of a solute in a solution in terms of amount of substance in a specified amount of mass of the solvent. Molar mass of KI = 39 + 127 = 166 g/mol.
20% aqueous solution of KI means 200 g of KI is present in 1000 g of solution. Therefore,  

Molality = Moles of KI / Mass of Water in kg

= (200/166) / (0.8) = 1.506 m

(b) Molarity is the concentration of a solution expressed as the number of moles of 
solute per litre of solution.
Given,
Density of the solution = 1.202 g/mL
Volume of 100 g solution = mass/ density
 = 100/1.202 
 = 83.19 mL
Therefore, molarity = 20/166 mol/ 83.19 * 10^-3 L 
 = 1.45 M
∴ Molarity of KI = 1.45M

(c) Molar mass of KI = 39 + 127 = 166 g/mol.
Moles of KI = 20/166 = 0.12 mol
Moles of water = 80/18 = 4.44 mol

Therefore,
Mole fraction of KI = Moles of KI / Moles of KI+ Moles of Water
 = 0.12 / 0.12+ 4.44
 = 0.0263
∴ Mole fraction of KI = 0.0263

16.28 

The commonly used soaps are the sodium and potassium salts of higher fatty acids. They are soluble in water and give good lather to it. On the other hand, calcium and magnesium salts of higher fatty acids (calcium and magnesium soaps) are insoluble in water and do not produce lather. Hard water contains Ca²⁺and Mg²⁺ ions. When a sodium or potassium soap is added to hard water, it gets converted into an insoluble calcium or magnesium soap as shown ahead.

Due to the conversion of sodium or potassium soaps into calcium or magnesium soaps in the presence of hard water, the common soaps are unable to emulsify the greasy dirt and clean the dirty object. They do not produce lather with hard water. The calcium and magnesium soaps thus produced appear on the surface as insoluble sticky grey scum. This involves a waste of the common soap and the fabrics being washed get discoloured and hardened. This is why the commonly used sodium and potassium soaps cannot be used as cleansing agents with hard water.

16.27 

It has been found that the detergents having a great deal of branching in the hydrocarbon tail are not biodegradable and cause pollution in rivers and waterways. This is because the presence of side chains in the hydrocarbon tail stops bacteria from attacking and breaking the chains.

This results in slow degradation of detergent molecules leading to their accumulation in water ways. Alkylbenzene sulphonates having branched chain alkyl groups possess poor biodegradability and cause pollution in rivers and water Such detergents are termed as non-biodegradable or hard detergents.

The alkyl sulphates do not possess branching and are therefore biodegradable. They do not cause pollution. Alkyl benzene sulphonates in which the phenyl group is either attached to the end of a long chain alkyl group are also biodegradable and do not cause pollution. These are termed as soft detergents.

Nowadays, efforts are being made to keep the amount of branching in the detergent molecule to a minimum. Since unbranched chains are more prone to attack by bacteria, the detergents having no branching or minimum branching are easily biodegraded and pollution is prevented. Example: Sodium-4- (1,3,5,7-Tetra Methyl Octyl) Benzene Sulphonate