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Candidates who are selected for admission must confirm their seat by paying the course fee. Here are few things that students will recieve after taking admission in Krishna Kanta Handiqui State Open University:

• Once a students get admission into KKHSOU, he or she will recieve the enrollment number and ID for future login to the samarth portal of the university.
• Based on the choice made by the candidate while filling the application form, he or she will then recieve the study material. Soft copy is available on the official website of the university. 
• Registered students also recieve a mail from the regional or study centre for counselling. It is an induction ceremony where the course is introduced to the students.
• A schedule of classes is also shared through mail. For any query regarding classes, students must contact their study centre.

let the molar masses of AB₂ and AB₄ be x and y respectively.

Molar mass of benzene (C₆H₆) = 12 × 6 + 1 × 6 = 78 g/mol

Moles of benzene = mass/molar mass = 20/78

n = 0.256mol

⇒ ΔT_f = 2.3 K

K_f = 5.1K kg mol^-1

For AB₂

Applying the formula: ΔT_f = K_f × M

Where

ΔT_f = depression in freezing point

K_f = molal depression constant

M = molality of solution

⇒ 2.3 = 5.1 × M₁

⇒ M₁ = 0.451mol/kg

For AB₄

Applying the formula: ΔT_f = K_f × M

ΔT_f = depression in freezing point

K_f = molal depression constant

M = molality of solution

⇒ 1.3 = 5.1 × M₂

⇒ M₂ = 0.255mol/kg

M₁ = moles of solute/mass of solvent (in kg)

M₁ = 1/x / 0.02 = 1 / 0.02x = 0.451

⇒ X = 110.86g

M₂ = moles of solute/mass of solvent (in kg)

M₂ = 1/y / 0.02 = 1 / 0.02y = 0.255

⇒ y = 196.1g

atomic mass of A be a and that of B be b g respectively.

So, AB₂ : a + 2b = 110.86

AB_4: a + 4b = 196.1

⇒ a = 25.59g

⇒ b = 42.64g

5% solution means 5g of cane sugar is present in 100g of solution

Freezing point of solution = 271k

Freezing point of pure water = 273.15k

Molar mass of cane sugar (C₁₂H₂₂O₁₁) = 12 × 12 + 1 × 22 + 16 × 11 = 342g

Moles of cane sugar = mass/molar mass = 5/342

⇒ n = 0.0146mol

Molality of solution = moles of solute/mass of solvent (in kg)

⇒ M = 0.0146/0.095

⇒ Molality = 0.154M

Depression in freezing point = ΔT_f = 273.15-271 = 2.15k

Applying the formula: ΔT_f = K_f × M

Where

ΔT_f = depression in freezing point

K_f = molal depression constant

M = molality of solution

⇒ K_f = 2.15/0.154

⇒ K_f = 13.96k kg mol^-1

Second condition: mass of glucose = 5g

Molar mass of glucose (C₆H₁₂O₆) = 12 × 6 + 1 × 12 + 16 × 6 = 180g

Moles of glucose = mass/molar mass

⇒ n = 5/180moles

⇒ n = 0.0278mol

Molality of solution = moles of solute/mass of solvent (in kg)

⇒ molality = 0.0278/0.095

⇒ M = 0.2926M

Applying the formula: ΔT_f = K_f × M

Where

ΔT_f = depression in freezing point

K_f = molal depression constant

M = molality of solution

⇒ ΔT_f = 13.96 × 0.2926

⇒ ΔT_f = 4.08k

Freezing point of solution = 273.15 + 4.08 = 277.234k

Given: mass of solute = 30g

Let the molar mass of solute be x g and vapour pressure of pure water at 298k be P₁ ^?

Mass of water(solvent) = 90g

Molar mass of water = H₂O = 1 × 2 + 16 = 18g

Moles of water = mass of water/molar mass

⇒ n = 90/18 moles

⇒ n = 5moles

Molar fraction of solute,

x₂ = moles of solute / moles of solute + moles of octane

x₂ = (30/x) / (30/x) + 5

x₂ = 30 / 30+5x

Vapour pressure of solution (p₁) = 2.8kpa

Applying the formula:

According to second condition when we add 18g of water to solution vapour pressure becomes 2.9kpa

Moles of water = mass/molar mass

⇒ n = 90 + 18/18

⇒ n = 6moles

Molar fraction of solute,

x₂ = moles of solute / moles of solute + moles of octane

x₂ = (30/x) / (30/x) + 6

x₂ = 30 / 30+6x

Applying the formula:

Solving 1 and 2:

Dividing (2) by (1) we get

Substituting value of x in 1 we get

(i) molar mass of the solute = 78g

(ii) vapour pressure of water at 298 K = 537kpa

As a state-open university, Krishna Kanta Handiqui State Open University offers distance/ part-time education among other open universities. Recognised by the Distance Education Bureau and the UGC, this open university in Uttarakhand offers more than 200 programmes at Certificate, Diploma, Degree, and Doctoral levels in Open and Distance Learning (ODL) and Online Learning (OL) modes. The University offers scholarships based on merit and need. In addition to that, KKHSOU also provides good placement opportunities along with assistance before and during placement drives. Therefore, it is considered one of the top universities to study at.

16.6

The basis on which drugs are classified in different ways is : - of drugs and the reasons for classification are as follows:

(i) On the basis of pharmacological effect:

This classification provides doctors the whole range of drugs available for the treatment of a particular type of problem. Hence, such a classification is very useful to doctors.

(ii) On the basis of drug action:

This classification is based on the action of a drug on a particular biochemical process.

(iii) On the basis of chemical structure:

This classification provides the range of drugs sharing common structural features and often having similar pharmacological activity.

(iv) On the basis of molecular targets:

This classification provides medicinal chemists the drugs having the same mechanism of action on targets. Hence, it is the most useful to medicinal chemists.