2.5 g of a non-volatile, non-electrolyte is dissolved in 100 g of water at 25°C. The solution showed a boiling point elevation by 2°C. Assuming the solute concentration is negligible with respect to t

Question

2.5 g of a non-volatile, non-electrolyte is dissolved in 100 g of water at 25°C. The solution showed a boiling point elevation by 2°C. Assuming the solute concentration is negligible with respect to the solvent concentration, the vapor pressure of the resulting aqueous solution is  mm of Hg (nearest integer)

[Given: Molal boiling point elevation constant of water , 1 atm pressure = 760 mm of Hg, molar mass of water = 18 g mol⁻¹]

Canvas Classes · Accepted Answer

Strategy:
 Relate the boiling point elevation (colligative property) to the molality, and then use the molality to find the mole fraction of the solvent, which gives the vapour pressure lowering via Raoult's Law.

Step 1: Calculate the molality ()
, .

Step 2: Determine the mole fraction of the solvent (water, )
In 100 g (0.1 kg) of water:
- Moles of solute 
- Moles of water

Step 3: Calculate the vapour pressure ()
Standard pressure at boiling point .

Note: Taking into account specific JEE

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