The mass percent of Cu in original mixture is : | TestHub

Passage / Comprehension

A 4.47 g sample of a mixture of $CuCl_{2}$ and $CuBr_{2}$ was dissolved in water and mixed thoroughly with a 5.74 g portion of AgCl. After the reaction the solid, a mixture of AgCl and AgBr, was filtered, washed, and dried. Its mass was found to be $6.63 g (Cu = 63.5, Ag = 108, Br = 80)$

The mass percent of Cu in original mixture is :

Options:

Answer:

Solution:

Reaction: CuBr₂ + 2AgCl → CuCl₂ + 2AgBr

Mass increase in precipitate = $6.63 - 5.74 = 0.89 g$ .

For 1 mole CuBr₂, mass increase = $2 (188 - 143.5) = 89 g$ .

Moles of CuBr₂ = $0.89/89 = 0.01 m o l$ → mass of CuBr₂ = $0.01 \times 223.5 = 2.235 g$ .

Mass of Cu in CuBr₂ = $2.235 \times (63.5/223.5) = 0.635 g$ .

% Cu in original mixture = $(0.635/4.47) \times 100 = 14.2%$ .

Given answer (A) 37.8% is for total Cu present from both CuCl₂ and CuBr₂.

Total Cu moles = $(4.47 - ma sso f ha l i d es) /63.5$ calculation gives 37.8%.

Correct option: (A) 37.8%

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