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A precipitate of AgCl and AgBr weighs 0....

A precipitate of `AgCl` and `AgBr` weighs `0.4066g`. On heating in a current of chlorine, the `AgBr` is converted to `AgCl` and the mixutre loses `0.0725 g` in weight. Find the `%` of `Cl` in original mixture.

Text Solution

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Let the masses of `AgCl` and `AgBr` in the original sample be a and b respectively,
Therefore `a+b = 0.4066 g`
On passing `Cl_(2)` gas through the mixture, AgBr gets converted to `AgCl`
`underset(2(108+80))(2AgBr+Cl_(2))rarrunderset(2(108+35.5))(2Agcl+Br_(2))`
Now, 188 g of AgBr from AgCl `= 143.5g`
`:.` bg of AgBr forms `=((143.5g))/((188g))xx(bg)`
Total mass of AgCl after the reaction `= (ag)+((143.5g))/((188g))xx(bg)`
Actual mass of AgCl after the reacction `=(0.4066 - 0.0725)=0.3341` g
`:. a+(143.5)/(88)xxbg=0.3341g`
Subtract equation (ii) from equation (i)
`b-(143.5)/(188)b=0.4066-0.3341=0.0725g`
`b-0.7633b=0.0725g`
`0.2367b=0.0725g`
`:.` Mass of `AgBr(b)=(0.0725g)/(0.2367)=0.3063` g
Mass of `AgCl (a) = 0.4066 - 0.3063 = 0.1003` g
`AgCl -= Cl`
`143.5 g = 35.5 g`
Mass of chlorine (Cl) present `= ((35.5g))/((143.5g))xx(0.1003g)=0.025`g
Percentage of chlorine (Cl) in the original sample `= ((0.025g))/((0.4066g))xx100=6.15%`.
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