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The quantity of Cl in a water supply is determined


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Answer #1

A) 20.2 ml = 0.0202 L

No. of mole = molarity \times volume of solution in liter

No. of mole of Ag+ = 0.1\times0.0202 = 0.00202 mole

According to reaction Ag+ and Cl- react in equimolar proportion thus, to to react 0.00202 mole of Ag+ 0.00202 mole of Cl- must be there

molar mass of Cl = 35.453gm/mole that mean 1 mole of Cl = 35.453 gm then 0.00202 mole Cl = 35.453\times0.00202 = 0.071615 gm of Cl

0.071615 gm of Cl in sample

B) 10 gm mass = 100%

then 0.071615 gm = 100\times0.071615/10 = 0.71615%

0.71615% of Cl- in solution.

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