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12 August, 18:52

A 0.3146-g sample of a mixture of nacl (s) and kbr (s) was dissolved in water. The resulting solution required 45.30 ml of 0.08765 m agno3 (aq) to precipitate the cl - (aq) and br - (aq) as agcl (s) and agbr (s). Calculate the mass percentage of nacl (s) in the mixture.

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  1. 12 August, 20:17
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    Let mass of NaCl in the mixture = x g

    Mass of KBr = 0.3146 - x

    Moles of NaCl = Mass / Molar mass = x / 58.5

    Mole sof KBr = 0.3146-x / 119

    Moles of AgNO3 added = molarity X volume (L) = 0.08765 X 45.30 / 1000 = 0.00397 moles

    Out of these say moles of AgNO3 reacted with Cl - = y

    So moles of AgNO3 reacted with Br - = 0.00397 - y

    We know that

    x / 58.5 = y ... (1)

    0.3146-x / 119 = 0.00397 - y ... (2)

    Putting (1) in (2)

    0.3146 - x / 119 = 0.00397 - x / 58.5

    18.40 - 58.5x = 27.64 - 119x

    x = 0.153 grams = Mass of NaCl

    MAss % of NaCl = Mass of NaCl X 100 / total mass

    = 0.153 X 100 / 0.3146 = 48.63%
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