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Grade 11Physical Chemistry

6.0g of a sample containing CuCl2 and CuBr2 is dissolved in 100 ml water. A 10 ml portion of this solution on treatment with AgNO3 solution results in complete precipitation of Cl- and Br- giving 0.9065 gram of precipitate. The precipitate thus obtained was shaken with dilute solution of NaBr where all AgCl is converted to AgBr. Mass of the new precipitate was found to be 1.005 g. Determine % mass of CuCl2 an CuBr2 in the original sample

Profile image of Harshit Chhikara
6 Years agoGrade 11
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2 Answers

Profile image of Vikas TU
6 Years ago
area is a good time to come by and look at it and see what I can do for you please let me know your thoughts and if so what time would definitely be in touch when I get to my life and my dad is going back home to
 
Profile image of Kshitij Bhosale
4 Years ago
6g mixture was dissolved in 100mL and 10mL was
taken
 ⇒ mass of mixture in 10 mL solution = 0.6 g.
 Suppose, in the 0.6g mixture:
 moles of CuCl2
 = x and moles of CuBr2
 = y
 \ moles of Cl-
 = 2x and moles of Br-
 = 2y
 \ moles of AgCl ppt. = 2x and moles AgBr ppt.=2y
 \ Adding masses of AgCl and AgBr:
 143.5 × 2x + 188 × 2y = 0.9065 (Eq(i))
 When the precipitate is shaken with NaBr, the AgCl
is also converted to AgBr.
 Thus, total moles of AgBr formed = (2x + 2y)
 \ total mass of ppt. = (2x + 2y) × 188
 \ 188 × 2x + 188 × 2y = 1.005 (Eq(ii))
 Eq(ii) - Eq(i) gives us
 44.5 × 2x = 0.0985
 ⇒ x = 1.106 × 10-3
\ Moles of CuCl2
 in 0.6g mixture = 1.106 × 10-3
 \ mass of CuCl2
 in 0.6 mixture = 0.15g
 \ % CuCl2
 in mixture = 0.6
0.15 × 100
 = 25% (Ans)
 Putting the value of x in Eq(i) we get:
 143.5 × 2 × 1.106 × 10-3
 + 188 × 2y = 0.9065
 ⇒ 376y = 0.9065 - 0.3174
 ⇒ y = 1.56 × 10-3
 \ Moles of CuBr2
 in 0.6g mixture = 1.57 × 10-3
 \ mass of CuBr2
 in 0.6g mixture
 = 1.57 × 10-3 × 188g
 = 0.2952g
 \ % of CuBr2
 in mixture = 0.6
0.2952 × 100
 = 49.2%

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