STANDARD Chemistry Colligative Properties MCQ Question
The van't Hoff factor of 0.005 M aqueous solution of KCl is 1.95. The degree of ionisation of KCl is
0.95
0.97
0.94
0.96
Correct Answer
Detailed Explanation
KCl dissociates into K⁺ and Cl⁻ ions. The degree of ionisation (α) can be calculated using the formula α = (i - 1) / (n - 1), where i is the van't Hoff factor and n is the number of ions formed. Substituting the values, α = (1.95 - 1) / (2 - 1) = 0.95.
Found an issue with this question?
Related Questions
More from Solutions
A 5% solution (by mass) of cane sugar in water has freezing point of 271 K and freezing point of pure water is 273.15 K. The freezing point of a 5% so...
Elevation in the boiling point for 1 molal solution of glucose is 2 K. The depression in the freezing point for 2 molal solution of glucose in the sam...
What amount of CaCl₂ (i = 2.47) is dissolved in 2 litres of water so that its osmotic pressure is 0.5 atm at 27°C?
More from
Calculate the total vapour pressure of an ideal solution at a certain temperature if the mole fraction of component 1 is 0.4, its vapour pressure in p...
A solution is made by mixing 200 g of ethanol (C2H5OH) with 300 g of water. If the vapor pressure of pure water at the same temperature is 24 mmHg and...