A solution containing 2.5 × 10–3 kg of a solute dissolved in 75 × 10–3 kg of water boils at 373.535 K. The molar mass of the solute is _____ g mol–1. [nearest integer] (Given : Kb(H2O) = 0.52 K kg mol–1 and boiling point of water = 373.15 K)
Correct Answer: 45
Solution and Explanation
The correct answer is 45
Wsolute = 2.5 × 10–3 kg
Wsolvent = 75 × 10–3 kg
ΔTb = 373.535 – 373.15= 0.385 K
Kb(H2O) = 0.52 K kg mol–1
\(△T_b=\frac{K_b×10^3×W_{solute}}{M_{solute}×W_{solvent}}\)
\(M_{solute}=\frac{0.52×10^3×2.5×10^{−3}}{75×10^{−3}×0.385}\)
= 45.02 ≈ 45
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Concepts Used:
Colligative Properties
Colligative Property of any substance is entirely dependent on the ratio of the number of solute particles to the total number of solvent particles but does not depend on the nature of particles. There are four colligative properties: vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure.
Examples of Colligative Properties
We can notice the colligative properties of arrangements by going through the accompanying examples:
- On the off chance that we add a spot of salt to a glass full of water, its freezing temperature is brought down impressively than its normal temperature. On the other hand, the boiling temperature is likewise increased and the arrangement will have a lower vapor pressure. There are also changes observed in its osmotic pressure.
- In the same way, if we add alcohol to water, the solution’s freezing point goes down below the normal temperature that is usually observed for either pure alcohol or water.
Types of Colligative Properties
- Freezing point depression: ΔTf =1000 x kf x m2 /(M2 x m1)
- Boiling point elevation: ΔTb = kb m
- Osmotic pressure: π = (n2/V) RT
- Relative lowering of vapor pressure: (Po - Ps)/Po