Ncert Solutions Chemistry Class 12th

Molarity = Moles of Solute / Volume of Solution in liter

(a) Given, In 4.3 L of solution there is 30 g of Co (NO₃)₂. 6H₂O

Molar mass of Co (NO₃)₂.6H₂O = (1 * 59 + 2 * (1 * 14 + 3 * 16) + 6 * 18)

= 291 g/mol.

∴ Moles = Given Mass / Molar Mass = 30/291 = 0.103 mol.

Now, Molarity = 0.103 mol / 4.3 L

= 0.023 M

(b) Given, 30 mL of 0.5 M H₂SO₄ diluted to500 mL.

In 1000 mL of 0.5 M H₂SO₄, number of moles present is 0.5 mol.

∴ In 30 mL of 0.5 M H₂SO₄, number of moles present = 30X 0.5 / 1000 mol.

= 0.015 mol.

∴ Molarity = 0.015 mol / 0.5L

= 0.03 M.

Let the total mass of the solution be 100 g and the mass of benzene be 30 g.

∴ Mass of carbon tetrachloride = (100 - 30) g = 70 g

Molar mass of benzene (C₆H₆) = (6 * 12 + 6 * 1) g mol ^-1

= 78 g mol ^-1

∴ Number of moles of C₆H₆ =30/78 mol

= 0.3846 mol

Molar mass of carbon tetrachloride (CCl₄) = 1 * 12 + 4 * 35.5

= 154 g mol ^-1

∴ Number of moles of CCl₄ = 70/154 mol

= 0.4545 mol

Thus, the mole fraction of C₆H₆ is given as:

Number of moles of C₆H₆ / Number of moles of C₆H₆  + Number of moles of CCl₄

= 0.3846 / (0.3846 + 0.4545)

Mass of Solution = Mass of Benzene + Mass of Carbon Tetrachloride

= 22 g + 122 g = 144 g

Mass percentage of Benzene = Mass of Benzene / Mass of Solution X 100 = 22/144 X 100 = 15.28%

Mass percentage of CCl₄ = Mass of CCl₄ / Mass of Solution X 100 = 122/144 X 100 = 84.72%

Given-

Mass of K₂SO₄, w = 25 mg = 25 X 10^-3 g,

Molar mass of K₂SO₄ = (39*2) + (32*1) + (16*4) = 174 g mol^-1

Volume V = 2 liter

T = 25⁰C + 273 = 298 K (add 273 to convert in Kelvin)

The reaction of dissociation of K₂SO₄ is written as,

K₂SO₄ → 2K ⁺ + SO₄^2-

Number if ions produced = 2 + 1 = 3, hence vant Hoff's factor, I = 3

Here, we use vant Hoff's equation for dilute solutions, given as,

^{πV = inRT}

where, n is the number of moles of solute, R is solution constant which is equal to the gas constant (0.082) and T is the absolute temperature (298 K).

Hence, the osmotic pressure of a solution is 5.27x10^-3atm

16.24

Alitame is a high potency artificial sweetener and it is not possible to control the sweetness imparted to food by alitame. Alitame is an artificial sweetener that is 2,000 times as sweet as sugar. This sweetness is very high as compared to the natural sugar and use of such sweetener is very critical while preparing sweet dishes. The structure of alitame is as follows:

Hence problem arises when alitame is used for sweetening purposes.

Given-

Vant Hoff's factor, I = 2.47

osmotic pressure, π = 0.75 atm

Volume of solution = 2.5L.

To determine the amount of CaCl₂, we use vant Hoff's equation for dilute solutions, given as,

πV = inRT

where, n is the number of moles of solute, R is solution constant which is equal to the gas constant and T is the absolute temperature.

Hence, the amount of CaCl₂ dissolved is 3.425g

Given-

K_H for O₂ = 3.30 * 10⁷ mm Hg,

K_H for N₂ = 6.51 * 10⁷ mm Hg

Percentage of oxygen (O₂) = 20 %

Percentage of nitrogen (N₂) = 79%

Total pressure = 10 atm

Using Henry's law,

where, p is the partial pressure of gas in the solution and K_H is Henry's constant.

Thus, the mole fraction of oxygen in solution, x_oxy = 4.61x10^-5

and the mole fraction of nitrogen in solution, x_nit is 9.22x10^-5

16.23 

Artificial sweetening agents like Saccharin, Alitame, Sucrolose can be used in the preparation of sweets for a diabetic patient as they do not add any calories to the body.

Diabetic people are advised to consume low-calorie diet [fewer carbohydrates & more proteinaceous and fiber rich]. The refined sugar like sucrose adds calorie to the diet but complex sugar and starches do not add to the calorie intake of a person and at the same time impart a sweet taste to diet. Artificial sweeteners are generally either complex sugar or protein in nature