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QJEEM18Chca9d: JEE Main 2018 Chemistry - Class 12 JEEM18Chca9d Solution: The mass of a non-volatile, non-electrolyte solute (molar mass…

The mass of a non-volatile, non-electrolyte solute (molar mass = 50 g mol-1 ) needed to be dissolved in 114 g octane to reduce its vapour pressure to 75%, is :

Multiple Choice Options

  • A 37.5 g
  • B 75 g
  • C 150 g
  • D 50 g

Step-by-Step Verified Solution

Molar mass of octane

(C8 H18) = 8 × 12 + 18 = 114 g/mol

Let, ω is the mass of solute.

Relative lowering vapour pressure,

ΔPP = ω50ω50+114114

⇒ 75100 = ω50ω50+1

⇒ 34 (ω50+1)=ω50

⇒ ω = 150 g

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