Activity Information 10Late Submission unta Today at 8 AM CST is alowed, with a

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Activity Information 10Late Submission unta Today at 8 AM CST is alowed, with a one time 10% penaty to ST bmitted score. Use the References to access important values if needed for this question A mixture consisting of only potassium bromide (KBr) and sodium bromide (NaBr) weighs 1.0451 g. When the mixture is dissolved in water and an excess of silver nitrate is added,all the bromide ions associated with the original mixture are precipitated as insoluble silver bromide (AgBr). The mass of the silver bromide is found to be 1.7822 g.Calculate the mass percentages of potassium bromide and sodium bromide in the original mixture. 9% Mass percent KBr Mass percent NaBr Submit Answer 5 question atempts remaining Nexdt Autosaved at 5:27 AM Back 6) 20p

Explanation / Answer

total weight of KBr + NaBr = 1.0451 or Wn + Wk = 1.0451 where Wn = weight of NaBr and Wk = weight of KBr

1 mole of NaBr contain 1 mole of Br

1 mole of KBr contain 1 mole of Br

number of moles of NaBr = Nn = Weight of NaBr/molecular weight of NaBr = Wn/102.894

similarly, number of moles of KBr = Weight of KBr/molecular weight of KBr = Wk/119.002

number of mole of NaBr + number of mole of KBr = total number of mole of Br in the mixture

let Nb be the number of moles of Bromine atoms.

therefore, (Wn/102.894) + (Wk/119.002) = Nb.................... (1)

as we know the weight of AgBr formed, we can calculate the number of moles of Br.

Here the number of moles of AgBr = number of moles of Br

number of moles of AgBr = 1.7822 g/187.77gmol-1

Nb = 9.4914 x 10-3

Substituting value of Nb in equation 1 we get

(Wn/102.894) + (Wk/119.002) = 0.0094914 .................... (2)

Equating (2) we get

119.002 Wn + 102.894 Wk = 116.2183 ................. (3)

We know total weight of KBr + NaBr = 1.0451

or Wn + Wk = 1.0451................... (4) (this is given in question)

multiply 4 by 119.002, we get 119.002 Wn + 119.002 Wk = 124.3690............. (5)

we have 119.002 Wn + 119.002 Wk = 124.3690............. (5)

119.002 Wn + 102.894 Wk = 116.2183 ................. (3)

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(5) - (3) = 16.108 Wk = 8.1507

ie Wk = 0.5060

as Wn + Wk = 1.0451, Wn = 0.5391

Mass percent of NaBr = Wn/1.0451 = 0.5391/1.0451 = 0.5158 = 51.58 %

Mass percent of KBr = Wk/1.0451 = 0.5060/1.0451 = 0.4842 = 48.42 %

Answer:

Mass percent of NaBr = 51.58 %

Mass percent of KBr = 48.42 %

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