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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. |
Answer» 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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