1.

The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non - voltaile solute in it.

Answer»

SOLUTION :1 molal solution means 1 mol of the solute is present in 1000 g of the solvent (WATER).
Molar mass of water = 18 g `mol^(-1)`
Therefore, Number of moles present in 1000 g of water `= (1000)/(18)=55.56` mol
Therefore, MOLE fraction of the solute in the solution is,
`x_(2)=(1)/(1+55.56)=0.0177`
It is given that,
Vapour presure of water `(p_(1)^(0))=12.3` kPa
Applying the relation,
`(p_(1)^(0)-p_(1))/(p_(1)^(0))=x_(2)=0.0177`
`x_(2)=(12.3-p_(1))/(12.3)=0.0177`
`RARR p_(1)=12.0823`
= 12.08 kPa (approximately)
Hence, the vapour pressure of the solution is 12.08 kPa.


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