Question

vapour pressure of ch3cl and ch2cl2 are 540 mm hg and 402 mm hg respectively. 101g of ch3cl and 85g of ch2cl2 are mixed together. determine:(i) the pressure at which the solution starts boiling. (ii) molar ratio of solute v/s solvent in vapour phase in equilibrium with solution.​

Answer 1

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Given:-

• Partial vapour pressure of CH3Cl is 540mmhg
• Partial vapour pressure of CH2Cl2 is 402mmhg
• 101g of CH3Cl and 85g CH2Cl2 is mixed

To find:-

• Vapour pressure kf mixture
• Molar ratio of Solute / solvent

Solution:-

moles of CH3Cl= 101/50.5 = 2moles

moles of CH2Cl2=85/85=1mole

Mole fraction of CH3Cl=2/3

Mole fraction of CH2Cl2=1/3

P total= P°aXa + P°bXb

P total= 540 x 2/3 + 402 x 1/3

P total= 360+134

P total=494mmhg

At 494 mm hg solution start boiling.

Now solute is CH2Cl2 and solvent is CHCl than there mole ratio is 2/1=2:1

Note:-

• P°a= partial vapour pressure of CH3Cl
• P°b=partial vapour pressure of CH2Cl2
• Xa=mole fraction of CH3Cl
• Xb=mole fraction of CH2Cl2