Chemistry - Mole Concept Question with Solution | TestHub

ChemistryMole ConceptCalculation of moleMedium2 minQB

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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 $6.63 g (Cu = 63.5, Ag = 108, Br = 80)$

% by mole of AgBr in dried precipate is :

Options:

Answer:

Solution:

Mass of dried precipitate = AgCl + AgBr = $6.63 g$ .

Increase in mass = $6.63 - 5.74 = 0.89 g$ (AgCl converted to AgBr).

Let the moles of AgBr formed be $x$ . The mass increase per mole of AgBr = $(188 - 143.5) = 44.5 g$ .

$x = 0.89/44.5 = 0.02 m o l A g B r$ .

Mass of AgBr = $0.02 \times 188 = 3.76 g$ , and AgCl mass = $6.63 - 3.76 = 2.87 g$ .

Mole % of AgBr = $\frac{0.02}{( 0.02 + 2.87/143.5 )} \times 100 = 50$ .

Stream:JEESubject:ChemistryTopic:Mole ConceptSubtopic:Calculation of mole

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