vapour pressure of pure water is 40mm.if a non volatile solute is added to it vapour pressure falls by 4mm,hence molarity of solution is

Dear ‚ÄčManideep Varma,

Po-PsPo=mole fraction of solutePo=V.P. of solventPs=V.P. of solutionmole fraction of solute=440=0.1so mole fraction of solute+mole fraction of solvent=1mole of solute=0.1mole of solvent=0.9wt of water present in 0.9 mole=0.9×18=16.2 g or 0.0162 kgmolality=moles of solutewt. of solvent in kg=0.10.0162=6.17molality of solution is 6.17.

Regards

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