Calculate the volume (in mL) of silver nitrate solution that will be needed for

Question

Calculate the volume (in mL) of silver nitrate solution that will be needed for the precipitation: normally the amount of silver nitrate required to precipitate all of the chloride in the sample is found by a trial and error process. However, to speed up the experiment, assume a sample mass of 0.1418 g, and that sample contains approximately 55.0% chloride ion by mass (some samples may contain more, some less), and calculate what volume of 0.100 M AgNO3(aq) needs to be added to precipitate all of the chloride ion. (Pay attention to the sig.figs. in your calculation; keep three sig.figs. in the answer.)

Explanation / Answer

mass of chloride in the sample = 0.1418 * 55% = 0.078 gm

Moles of chloride = 0.078 gm/35.5g/mol = 2.196 *10^-3 moles

Suppose V litre of AgNO3 is required.

V * 0.1M = 2.196 *10^-3

or, V = 0.2196 * 10^-3 L = 0.220 mL

volume of AgNO3 required to precipitate all the chloride = 0.220 mL

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