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QJEEM19Ch6a9d: JEE Main 2019 Chemistry - Class 12 JEEM19Ch6a9d Solution: Freezing point of a 4% aqueous solution of X is equal to…

Freezing point of a 4% aqueous solution of X is equal to freezing point of 12% aqueous solution of Y. If molecular weight of X is A, then molecular weight of Y is -

Multiple Choice Options

  • A 4A
  • B 2A
  • C 3A
  • D A

Step-by-Step Verified Solution

For same freezing point,

(ΔTf)X = (ΔTf)Y

⇒ kf mx = kf my

⇒ 4×1000A×96=12×1000M×88

⇒ M = 3.27A ≃ 3A

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