Question

A 0.2 ℅ of solution of a non - volatile solute exerts vapour pressure of 1.004 bar at 100°C
What is the molar mass of the solute​

Answer 1

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HERE IS YOUR ANSWER:-)

#=>According to Raoult's law,

P0−PsP0=P0−PsP0=nB=nBnA+nBnB=nBnA+nB

⇒nBnA⇒nBnA

⇒WB×MAMB×WA⇒WB×MAMB×WA

(For dilute solutions,nB<<<k)

⇒0.2×1899.8×1.0131.013−1.009⇒0.2×1899.8×1.0131.013−1.009

⇒4.06gmol−1

HOPE IT HELPS YOU..

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Answer 2

According to Raoult's law

p°-p/p°=nB(mole fraction of solute )

=nB/nA+nB

(For dilute solution, nB<<<nA)

Mb=Eb×Ma/Wa. p°/p°-p

=0.2×18/99.8×1.013/1.013-1.009

=4.06gmol-1