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ADNAN MUHAMMED

Grade 12,

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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

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

Grade:11

2 Answers

Vikas TU
14149 Points
3 years ago
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Kshitij Bhosale
13 Points
2 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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