In an experiment to determine solubility of sodium chloride, 10.0 cm³ of a saturated solution of sodium chloride weighing 10.70g were placed in a volumetric...

In an experiment to determine solubility of sodium chloride, 10.0 cm³ of a saturated solution of sodium chloride weighing 10.70g were placed in a volumetric flask and diluted to a total of 500cm³. 25.0 cm³ of the diluted solution of sodium chloride reacted completely with 24.0 cm³ of 0.1M silver nitrate solution.

I. Calculate;
(i) Moles of silver nitrate in 24.0 cm³ of solution.
(ii) Moles of NaCl in 25.0 cm³ of solution.
(iii) Moles of NaCl in 500 cm³ of solution.
(iv) Mass of NaCl in 10.0 cm³ of saturated sodium chloride (Na = 23, Cl = 35.5)
(v) Mass of water in 10.0cm³ of saturated solution.
(vi) The solubility of NaCl in g/100g of waters.

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25 cm³
= 0.048 moles

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maurice.mutuku answered the question on October 25, 2019 at 06:35

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