Solutions

(i) Velocity of reaction.
The velocity (rate) of a chemical reaction is the change in concentration of reactants or products per unit time. (ii) Raoult’s law.
According to Raoult’s law, for a solution of non-volatile solute: where , = vapour pressure of pure solvent, = mole fraction of solute.
Step 3: Conclusion.
- Reaction velocity = rate of change of concentration.
- Raoult’s law = vapour pressure lowering proportional to solute mole fraction.

Given:
10% (w/w) solution 10 g glucose in 100 g solution.
Mass of solution = 100 g.
Mass of solute (glucose) = 10 g.
Mass of solvent (water) = 90 g = 0.090 kg.
Molar mass of glucose = 180 g/mol.
Density = 1.20 g/mL = 1200 g/L.
(i) Molality:
(ii) Molarity:
(iii) Mole Fraction:
Final Answer:
Molality = 0.617 mol/kg
Molarity = 0.667 M
Mole fraction of glucose = 0.011
Mole fraction of water = 0.989

Step 1: Definition.
Osmotic pressure is the excess pressure applied to a solution to prevent the inward flow of solvent molecules across a semipermeable membrane. Step 2: Expression for osmotic pressure.
where, = osmotic pressure, = molar concentration of solute, = gas constant, = absolute temperature. Step 3: Proof as a colligative property.
From the equation, osmotic pressure depends only on the number of solute particles ( ) in a given volume of solution, and not on their nature. Therefore, it is a colligative property. Conclusion:
Osmotic pressure is directly proportional to solute concentration and depends only on the number of solute particles, hence it is a colligative property.

Step 1: Recall dependence of concentration units.
- Molarity (moles of solute per litre of solution) depends on volume, and since volume changes with temperature, molarity changes with temperature.
- Normality is also based on volume, so it changes with temperature.
- Mass-volume percentage depends on solution volume, hence temperature dependent.
- Molality (moles of solute per kg of solvent) depends on mass, and mass does not change with temperature. Thus, molality is independent of temperature.

Step 2: Conclusion.
Therefore, the concentration unit that is independent of temperature is molality.

Final Answer:

Step 1: Formula for Elevation in Boiling Point.
The elevation in boiling point is given by the formula: where: - is the change in boiling point, - is the ebullioscopic constant of the solvent (2.50 K kg mol ), - is the molality of the solution. Step 2: Calculate the Change in Boiling Point.
Given that the boiling point of the solution is 350.50 K and the boiling point of the pure liquid is 350 K: Step 3: Calculate the Molality.
Rearranging the formula to solve for molality: Step 4: Calculate the Moles of Solute.
Molality is also defined as: The mass of the solvent is 100 g = 0.1 kg. Substituting the values: Step 5: Calculate the Molar Mass of the Solute.
The molar mass is given by:

Step 1: Formula for molality.
The molality is given by: Step 2: Calculate moles of acetic acid.
Moles of acetic acid = Step 3: Calculate molality.
Mass of benzene = 80 g = 0.080 kg Molality = Step 4: Conclusion.
Hence, the correct answer is (B) 0.00625 mol kg .

Step 1: Definitions.
- Osmosis: Flow of solvent molecules through a semipermeable membrane from lower solute concentration to higher solute concentration.
- Osmotic pressure: The pressure required to stop osmosis.
Step 2: Moles of protein.
Step 3: Concentration of solution.
Volume = 200 cm = 0.200 L.
Step 4: Osmotic pressure formula.
Step 5: Conclusion.
Osmotic pressure = .

Step 1: Mass percent formula.
Mass percent = .
Step 2: Calculation.
Step 3: Conclusion.
Therefore, glucose present =
20 g.

Step 1: Understanding the problem.
We are given a 30% by weight aqueous solution of glycerol. This means that in 100 g of solution, glycerol contributes 30 g and water contributes 70 g.
Step 2: Calculate moles of glycerol.
Molecular mass of glycerol (C H O ) = (3 × 12) + (8 × 1) + (3 × 16)
= 36 + 8 + 48
= 92 g/mol.
Moles of glycerol = mol.
Step 3: Calculate moles of water.
Molecular mass of water (H O) = 18 g/mol.
Mass of water = 70 g.
Moles of water = mol.
Step 4: Calculate mole fraction of glycerol.
Total moles = 0.326 + 3.889 = 4.215 mol.
Mole fraction of glycerol = .
Conclusion:
The mole fraction of glycerol in 30% by weight aqueous glycerol solution is approximately:

Step 1: Recall formula.
where = molarity, , . Step 2: Calculate molarity.
5% (w/v) = 5 g urea in 100 mL solution = 50 g in 1 L. Moles of urea = . Molarity = 0.833 M. Step 3: Substitution.
Step 4: Conclusion.
Osmotic pressure = .

Step 1: Recall osmosis.
In osmosis, solvent flows naturally from dilute to concentrated solution across a semipermeable membrane.
Step 2: Applying pressure.
If pressure greater than osmotic pressure is applied on the concentrated side, the natural flow is reversed. Solvent moves from concentrated solution to dilute solution. This is called anti-osmosis or reverse osmosis.
Step 3: Example.
Reverse osmosis is used in water purification, such as desalination of seawater where pure water passes through the membrane leaving salts behind.
Step 4: Conclusion.
Thus, anti-osmosis is reverse osmosis caused by applying external pressure greater than osmotic pressure.

Step 1: Definition of molarity.
Molarity (M) = .

Step 2: Find moles of water.
- Molar mass of water = 18 g mol .
- Density of water = 1000 g L .
So, number of moles in 1 L of water = mol.

Step 3: Conclusion.
Thus, the molarity of pure water is 55.56 mol L .

Final Answer:

Step 1: Definition of Osmotic Pressure.
Osmotic pressure is the pressure required to stop the flow of solvent molecules through a semipermeable membrane from pure solvent to solution. It is a colligative property and depends on the number of solute particles in solution, not their nature.
Step 2: Mathematical Expression.
The osmotic pressure is given by: where, - = osmotic pressure
- = molar concentration of solute
- = gas constant
- = absolute temperature Step 3: Relation with Molar Mass.
Molar concentration can be written as: where, - = mass of solute
- = molar mass of solute
- = volume of solution (in litres) Substituting in the osmotic pressure formula: Conclusion:
The relation between osmotic pressure and molar mass of solute is:

Step 1: Recall osmotic pressure formula.
where = van’t Hoff factor.
Step 2: For urea.
Urea is non-electrolyte → .
Step 3: For NaCl.
NaCl dissociates completely → . So effective concentration doubles.
Step 4: Conclusion.
Osmotic pressure of NaCl solution is higher than that of urea.