Multiple choice

A mixture of $AgCl$ and $AgBr$ undergoes a loss in wt. by $5$% if it is exposed to excess chlorination. The % of $AgCl$ in the mixture.

  1. $21.12$%
  2. $78.88$%
  3. $5$%
  4. $95$%
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B Correct answer
Explanation

Let x be the mass of AgCl and y be the mass of AgBr. The molar masses are approximately AgCl=143.3 and AgBr=187.8. Chlorination converts AgBr to AgCl, resulting in a mass loss of (187.8 - 143.3) = 44.5 per mole of AgBr. The 5% loss implies 0.05(x + y) = (44.5/187.8)y. Solving this ratio leads to the percentage of AgCl.

AI explanation

Let the mass of silver chloride be x grams and the mass of silver bromide be y grams. The mass of silver bromide converts to silver chloride, replacing a bromine atom of mass 80 with a chlorine atom of mass 35.5, which causes the measured 5 percent mass increase relative to the initial total mass of the mixture. Solving the equation 44.5 multiplied by y divided by 187.5 equals 0.05 multiplied by x plus y gives y equal to 21.12 percent of the mixture and x equal to 78.88 percent. The percentage of silver chloride is 78.88 percent.

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