Chemistry

16.22 What are artificial sweetening agents? Give two examples.

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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.

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Q 2.37 Vapour pressures of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot ptotal, pchloroform, and pacetone as a function of xacetone. The experimental data observed for different compositions of mixture is: Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal solution.

100 xacetone	0	3.4.8	23.4	36.0	50.8	58.2	64.5	72.1
pacetone /mm Hg	0	54.9	110.1	202.4	322.7	405.9	454.1	521.1
pchloroform /mm Hg	632.8	548.1	469.4	359.7	257.7	193.6	161.2	120.7

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

ptotal (mm Hg)

632.8

603.0

579.5

562.1

580.4

599.5

615.3

641.8

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.

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16.21 Why is use of aspartame limited to cold foods and drinks?

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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:

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Q 2.36 100 g of liquid A (molar mass 140 g mol^–1) was dissolved in 1000 g of liquid B (molar mass 180 g mol^–1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.

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Given-

Mass of liquid A, W_A = 100g, Molar mass, M_A = 140 g mol^-1

Mass of liquid B, W_B = 1000 g, Molar mass, M_B = 180 g mol^-1

Using the formula below calculate the no. of moles in liquid A and B.

Number of moles = Mass / Molar Mass

Number of moles of liquid A, M_A = 100/140 = 0.714 mol^-1

Number of moles of liquid B, M_B = 1000/ 180 = 5.556 mol^-1

Using the formula,

mole fraction of a liquid = No. of moles of the liquid / total no of moles

we calculate the mole fraction of liquids A and B.

→ Mole fraction of A,

x_A = 0.714 / (0.714 + 5.556)

∴ x_A= 0.114

→ Mole fraction of B,

x_B = 1- x_{A = 1 - 0.114}

∴ x_B= 0.886

Vapour pressure of pure liquid B, P^o

...more

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16.20 What are food preservatives?

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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.

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Q 2.35 Henry's law constant for the molality of methane in benzene at 298 K is 4.27 * 10^5 mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

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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

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Q 2.34 Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.

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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

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Q 2.33 19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.00 C. Calculate the van't Hoff factor and dissociation constant of fluoroacetic acid.

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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

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16.19 What is tincture of iodine? What is its use?

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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.

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Q 2.32 Calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250 g of water. Ka = 1.4 * 10^–3 , Kf = 1.86 K kg mol^–1 .

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