Plus Two Chemistry Chapter Wise Previous Questions Chapter 2 Solutions is part of Kerala Plus Two Chemistry Chapter Wise Previous Year Questions and Answers. Here we have given Plus Two Chemistry Chapter Wise Questions and Answers Chapter 15 Polymers.
Kerala Plus Two Chemistry Chapter Wise Previous Questions and Answers Chapter 2 Solutions
A solution contains 15 g urea (molar mass=60 g mot-1) per litre of solution in water has the same osmotic pressure as a solution of glucose (molar mass= 180 g mol-1) in water. Calculate the mass of glucose present in one litre of its solution. [March 2018]
Here the two solutions have the same osmotic pressure. So they are isotonic.
We know that
Since , it follows that (at constant temperature and volume)
or, (Here 1 indicates urea and 2 indicates glucose)
Mass of glucose = 45g f
An element crystallises as FCC with density 2.8 g cm-3. It’s a unit cell having edge length 4 × 10-8 cm. Calculate the molar mass of the element. (Given NA=6.022 × 1023 mol-1). [March 2018]
Here d = 2.8g/cm3, a = 4 x 10-8 cm, z = 4 ( since fee) and NA = 6.022 × 1023 We know that
a. Henry’s law is related to the solubility of a gas in the liquid.
i. State Henry’s law
ii. Write any two application of Henry’s law.
b. 1000 cm3 of an aqueous solution of a protein contains 1,26 gm of the protein. The osmotic pressure of such a solution at 300K is found to be 2.57 × 10-3 bar. Calculate the molar mass of protein. (R = 0.083 L bar mol-1 K-1) [March 2017]
i. It states that “the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of liquid or solution”. The most commonly used form of Henry’s law states that “the partial pressure (p) of a gas in the vapour phase is proportional to the mole fraction of the gas (x) in the solution” and it is expressed as, where, is Henry’s law constant and ‘x’ is mole fraction of gas in solution.
1. To increase the solubility of CO2 in soda water and soft drinks, the bottle is sealed under high pressure.
2. To avoid toxic effects of high concentration of nitrogen in blood, the tanks used by scuba divers are filled with air diluted with helium,
b. p = 2.57 × 103 bar
V = 1000 cm3 = 1L T = 300K
R = 0.083 L bar mol-1 K-1
i. Number of moles of the solute per kilogram of the solvent is
a. Mole fraction
c. Molarity d. Molar mass
ii. ‘The extent to which a solute is dissociated or associated can be expressed by Can’t Hoff factor.’ Substantiate the statement. ‘
iii. The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, nonelectrolyte solid weighing 0.5 g when added to 39 g of benzene (molar mass 78 g/mol), vapour pressure becomes 0.845 bar. What is the molar mass of the solid substance? [March 2016]
i. b. Molality
ii. In the case of dissociation, the observed molecular weight has a lower value than the normal molecular weight while in the case of an association, the observed molecular weight is higher than the normal molecular weight. In order to explain such abnormal cases, Can’t Hoff factor is used.
This factor is defined as.
i. Among the following which is not a colligative property?
a. Osmotic pressure
b. Elevation of boiling point
c. Vapour pressure
d. Depression of freezing point
ii. a. 200 cm3 of aqueous solution of a protein contains 1.26 g of protein. The osmotic pressure of solution of 300 K is found to be 8.3 × 10-2 bar. Calculate the molar mass of protein. (R = 0.083 Lbar k-1 mol-1)
b. What is the significance of Van’t Hoff factor? [March 2015]
i. Vapour pressure
ii. a. Here π = 8.3 × 10-2 bar,
b. In the case of dissociation, the observed molecular weight has a lower value than the normal molecular weight while in the case of an association, the observed molecular weight is higher than the normal molecular weight. In order to explain such abnormal cases, Can’t Hoff factor is used. This factor is defined as.
Osmotic pressure is a colligative property and it is proportional to the molarity of solution.
a. What is osmotic pressure?
b. Molecular mass of NaCI determined by osmotic pressure measurement is found to be half of the actual value. Account for it.
c. Calculate the osmotic pressure exerted by a solution prepared by dissolving 1.5g of a polymer of molar mass 185000 in 500 ml of water at 37 °C. [R = 0.0821 L atm/K/ mol] [March 2014]
a. Osmotic pressure is the external pressure required to stop the flow of solvent molecule from higher concentration to lower concentration through the semipermeable membrane. It is a colligative property.
b. On dilution number of molecules increases, therefore mass decreases. On dilution, NaCI dissociates to 2 molecules. So molar mass decreases to half.
v-volume of solvent
W2 -weight of solute,
M2-Molar mass of solute, R universal gas constant,
Colligative properties are properties of dilute solutions which depend on the number of solute particles.
a. Name the four important colligative properties.
b. Which colligative property can provide the molar mass of proteins with greater precision?
c. Calculate the temperature at which a solution containing 54g glucose in 250g of water will freeze. (K1 for water = 1.86 K/mol kg). [Model 2014]
a. 1. Lowering of vapour pressure
2. Elevation of boiling point
3. Depression of freezing point
4. Osmotic pressure
b. Osmotic pressure
Elevation of boiling point is a colligative
a. What are colligative properties?
b. Elevation of boiling point (ΔTb) is directly proportional to molality (m) of a solution. Thus ΔTb = Kbm ; Kb is called the molal elevation constant.
From the above relation derive an expression to obtain molar mass of the solute.
c. The boiling point of benzene is 353.23K. When 1.80 g of a nonvolatile solute was dissolved in 90 g of. benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. (Kb for benzene is 2.53 K kg mol-1). [March 2013]
a. Properties which depend on the number of particles and not on its nature are called colligative properties. The mixture is known as a minimum boiling azeotrope.
If W1 = weight of the solvent
W2 = weight of the solute
M2 = molar mass of the solute
m = molality of the solution, then,
Vapour pressure of a solution is different from that of the pure solvent.
a. Name the law which helps us to determine partial vapour pressure of a volatile component in solution.
b. State the above law.
c. Vapour pressure of chloroform (CHCI3) and dichloromethane (CH2CI2) at 298 K are 200 mm and 415 mm of Hg respectively. Calculate the vapour pressure of solution prepared by mixing 24 g of chloroform and 17 g of dichloromethane at 298 K. [Atomic mass: H: 1, C: 12, Cl: 35.5] [March 2012]
b. For a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction.
c. Molar mass of CH2CI2 = 12 x 1 + 1 x 2 + 35.5 x 2 = 85 g/mol
Molar mass of CHCI3= 12 x 1 + 1 x 1 + 35.5 x 3 = 119.5 g/mol
No. of moles of CH2CI2 moles
No. of moles of CHCL3 = moles
Total no. of moles = 0.2 + 0.2 = 0.4 moles
Mole fraction of CH2CI2,
Vapour pressure of the solution,
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