The addition of 3 gms of a substance to 100 gms `C Cl_(4)` (M.w. = 154 gm/mole) raises the boiling point of `C Cl_(4)` by `0.60^(@)C`. If `K_(b)` of `C Cl_(4)` is 5.03 K Kg `"mole"^(-1)` then find out the relative lowering of vapour pressure.

Options

A. `0.0181`
B. `0.0224`
C. `0.204`
D. `0.192`

Answer

A

Explanation

Substance = 3 gms `= W_(2)`
Mol. Wt. of sub `= M_(2)=?`
Solvent `= C Cl_(4)=100 gms = W_(1)`
Molal of solvent `C Cl_(4)=153" gm mole"^(-1)=M_(1)`
Increase of B.P. `= Delta T_(b)=0.6^(@)C`
`K_(b)` of solvent `C Cl_(4)=5.03" K kg mole"^(-1)`
`M_(2)=(K_(b)xx W_(2)xx 100)/(Delta T_(b)xx W_(1))`
`therefore M_(2) = (5.03xx3xx1000)/(0.6xx100)`
`= 251.5"gm mol"^(-1)`
Relative depression of vapour pressure
`= (p^(@)-p)/(p^(@))=X_(2)`
`= (n_(2))/(n_(1)+n_(2))`
Where `n_(2)=` mole of solvent
`= (W_(2))/(M_(2))=(3)/(251.5)`
= 0.0119 mole substance
`n_(1)=` mole of solvent
`= (W_(1))/(M_(1))=(100)/(154)`
= 0.6494 mole solvent `C Cl_(4)`
Total mole `= 0.6494+0.0119=0.6613`
`X_(2)=(0.0119)/(0.6613)`
`=0.01799 ~~0.0181`