Solutions

Step 1: Osmotic pressure is given by the relation where is osmotic pressure, is molarity, is gas constant, and is absolute temperature.
Step 2: Given:
Step 3: Rearranging the formula,
Step 4: Substituting values, Hence, the required molarity is

Step 1: Calculate moles of oxygen liberated at STP:
Step 2: Decomposition reaction: From the equation, 1 mol of is produced from 2 mol of .
Step 3: Mass of :
Step 4: Percentage strength by weight: Hence, the percentage strength of the solution is

Step 1: Elevation in boiling point is given by
Step 2: Substituting given values,
Step 3: Molality means 1 mole of urea is dissolved in 1 kg (1000 g) of water. Moles of water:
Step 4: Mole fraction of urea:
Step 5: Rounding off to two decimal places, Hence, the mole fraction of urea is

Step 1: According to Henry’s law, at constant temperature, the solubility of a gas in a liquid is directly proportional to its partial pressure.
Step 2: Hence, Given:
Step 3: Calculate the new solubility:
Step 4: Therefore, the solubility of nitrogen at pressure is:

Step 1: Recall the role of Van’t Hoff factor. The Van’t Hoff factor accounts for the number of effective particles in solution. All colligative properties depend directly on .
Step 2: Write the relation for elevation in boiling point. Thus,
Step 3: Compare the given Van’t Hoff factors. Greater the value of , greater is the elevation in boiling point.
Step 4: Arrange the boiling points. Hence, Therefore, the correct option is

Step 1: Understand the meaning of isotonic solutions. Two solutions are isotonic if they have the same osmotic pressure at the same temperature. Step 2: Calculate molarity of the urea solution. A solution means: So, in : Molar mass of urea Step 3: Compare van’t Hoff factors. Urea is a non-electrolyte: Glucose is also a non-electrolyte: Step 4: Apply isotonic condition. For isotonic solutions: Hence, a solution of urea is isotonic with a solution of glucose.

Step 1: Identify the property of calcium chloride. Calcium chloride is a highly soluble salt and acts as a strong electrolyte. When dissolved in water, it dissociates into ions: Step 2: Effect on freezing point. The presence of ions increases the number of particles in solution, which causes a lowering of freezing point (freezing point depression). Step 3: Application on roads in polar regions. When is sprinkled on snowy or icy roads:

It lowers the freezing point of water,
Prevents water from freezing easily,
Helps in melting existing snow and ice.
Step 4: Final conclusion. Thus, calcium chloride is used on roads to minimise snow accumulation.