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Question

What is the mass of the precipitate formed when 50mL of 16.9% (w/v) solution of AgNO3 is mixed with 50mL of 5.8% (w/v) NaCl solution?
[Ag=107.8,N=14,O=16,Na=23,Cl=35.5]

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Solution

Moles of AgNO3=50×16.9100×169.8=0.05 mole

Moles of NaCl=50×5.8100×58.5=0.05 mole

AgNO3+NaClAgCl+NaNO3

Mass of AgCl precipitate= Mole × Molar mass

=0.05×143.5

=7.16g

Hence, 7.16 g AgCl will be precipitated out.

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