The vapour pressure of C Cl_(4)" at "30^(@)C is 143 mm of Hg.0.5 gm of a nonvolatile non electrolyte substance with molar mass 65 is dissolved in 100 ml of C Cl_(4). What will be the vapour pressure of the solution. Density of C Cl_(4)" at "30^(@)C = "1.58 gm per cc."

The vapour pressure of C Cl_(4)" at "30^(@)C is 143 mm of Hg.0.5 gm of

1.	The vapour pressure of C Cl_(4)" at "30^(@)C is 143 mm of Hg.0.5 gm of a nonvolatile non electrolyte substance with molar mass 65 is dissolved in 100 ml of C Cl_(4). What will be the vapour pressure of the solution. Density of C Cl_(4)" at "30^(@)C = "1.58 gm per cc."
Answer» Solution :Vapour pressure of pure solvent, `P^(@)=143`mm of HG. Vapour pressure of solution, `p = ?` Weight of solute, `W_(2) = 0.5 gm` Mol. wt of solute, `M_(2) = 65` Mol. wt of solvent `(C Cl_(4)), M_(1) = 154` Weight of solvent, `W_(1) = 100xx1.58 =158 gm` `("MASS" =" DENSITY"xx "volume")` `(p^(@)-p)/(p^(@))=(W_(2))/(M_(2)).(M_(1))/(W_(1))"(By Raoult.s law)"` `therefore (143-p)/(143)=(0.5)/(65)xx(154)/(158)` `therefore"p = 141.93 mm of Hg."`