Ncert Solutions Chemistry Class 12th

Given-

Thus, mole fraction of benzene in vapour phase is 0.6744

16.22 

Artificial sweetening agents refer to those compounds which impart sweet taste to any food product but at the same, they do not add any calories to the body. Example: Aspartame, Alitame, Saccharin etc.

The P_total for the values given in the graph is found out and plotted in the graph.

It can be observed from the graph that the plot for the p total of the solution curves downwards. Therefore, the solution shows negative deviation from the ideal behaviour.

16.21 

At elevated temperatures, aspartame is unstable and break down to give a tasteless compound because of which aspartame is limited to cold foods and drinks.

Aspartame is methyl ester of the dipeptide obtained from phenylalanine and aspartic acid. It is about 180 times as sweet as cane sugar. The structure of aspartame is as follows:

16.20 

Food preservatives are chemicals that prevent food from spoilage due to microbial growth. Table salt, sugar, vegetable oil, sodium benzoate (C₆H₃COONa), and salts of propanoic acids are some common examples of food preservatives.

Given-

Henry's law constant K_H = 4.27X 10⁵ mm Hg,

p = 760mm Hg,

Using Henry's law,

Using the formula of lowering vapour pressure,

Thus, the solubility of methane in benzene is 0.023 moles

Given- Vapour pressure of water,

P_A⁰  = 17.535 mm Hg  

W_B= 25 g of glucose

W_A = 450g of water

Molar mass of water, H₂O = 1 + 1 + 16 = 18 g mol^-1

Molar mass of glucose, C₆H₁₂O₆ = (12*6) + (1*12) + (16*6) = 180 g mol^-1

Using Raoult's law for solution of non-volatile solute,

P_A⁰ - P_A / P_A⁰ = x_B? Equation 1

where x_B is the mole fraction of the solute

x_B = W_B/M_B X M_B/W_B

=25/180 X 18 / 450

x_B = 1/180

Substituting the value of x_B in equation 1, we get,

Thus, the vapour pressure of water at 293 K at the given conditions is 17.437 mm Hg

Given- w₁ = 500g

W₂ = 19.5g

K_f = 1.86 K kg mol^-1

Molar mass of CH₂FCOOH = 12 + 2 + 19 + 12 + 16 + 16 + 1

= 78 g mol^-1

The depression in freezing point is calculated by,

→ (where, m is the molality)

= 1.86 X 19.5 / 78 X 1000/500

= 1.86 X 19.5 / 78 X 2

=0.93

∴ Δt_f (calculated) = 0.93

To find out the vant Hoff's factor, we use the formula,

i = observed Δt_f / calculated Δt_f

_{i = 1.0 (given) / 0.93}

∴ i= 1.07

CH₂FCOOH → CH₂FCOO^- + H ⁺

To find out the degree of dissociation α, we use

Thus, the vant Hoff's factor is 1.07 an the dissociation constant is 2.634x10^-3

16.19

Iodine is a powerful antiseptic. It is employed as tincture of iodine which is an alcohol-water solution containing 2-3 percent of iodine. Iodine is widely used in healing and treatment of wounds as it kills all the microbes present in the wounded region.