The vapour pressure of pure `C""Cl_(4)` (molar mass `=154gmol^(-1)`) and `SnCl_(4)` (molar mass `=170gmol^(-1)` ) at `25^(@)C` are 115.0 and 238.0 torr respectively. Assuming ideal behaviour, calculate the total approximate vapour pressure in torr of a solution containing 10 g of `C""Cl_(4)` and 15g of `SnCl_(4)`.

Options

A. 185.85
B. `190.0`
C. 180.7
D. 182.1

Answer

A

Explanation

Given , vapour pressure of pure ` C Cl_(4) "at " 25^(@) C `
`(p^(@)""_(SnCl_(4))) = 1150 ` torr
Vapour pressure of pure ` SnCl_(4) " at " 25^(@) C `
` (p^(@) ""_(SnCl_(4)) = 238.0` torr
Molar mass of ` C Cl_(4) [M_(C Cl_(4))] =154 " g mol" ^(-1)`
Molar mass of ` SnCl_(4) [M_(SnCl_(4))] = 170 " g mol"^(-1)`
Mass of ` C Cl_(4)` in solution ` (W_(C Cl_(4))) = 10g `
Mass of ` SnCl_(4)` in solution ` (W_(SnCl_(4))) = 15 g `
From Dalton's law of partial pressurre,
` P_("total") = chi_(C Cl_(4)) .P^(@)""_( C Cl_(4)) + chi_(SnCl_(4)) . P^(@) SnCl_(4))`
and ` chi = (n_("simple"))/(n_("total") ) = (w)/(Mxxn_("total")) `
Now , ` n = [w/M]_(C Cl_(4) ) = (10)/(154) = 0.065`
and ` [w/M]_(SnCl_(4) = (15)/(170) = 0.09`
Thus , ` chi_(C Cl_(4) )= n_(C Cl_(4))/(n_(C Cl_(4)) + n_(SnCl_(4)))`
` = (0.065)/(0.06 + 0.09)`
`chi _(C Cl_(4)) = (0.065)/(0.15) = 0.43 `
Therefore , ` chi_(SnCl_(4)) = 1 - 0.43 = 0.57 `
Thus , ` P_("total" ) = 0.43 xx 11500 + 0.57 xx 238`
` = 185.11 ~~ 18585 ` torr
Hence , option (a) is the correct answer .