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expeople1 [14]
3 years ago
13

If 26.2 mL of AgNO3 is needed to precipitate all the Cl− ions in a 0.785-mg sample of KCl (forming AgCl), what is the molarity o

f the AgNO3 solution?
Chemistry
1 answer:
sweet-ann [11.9K]3 years ago
6 0

<u>Answer:</u>

<em>The molarity of the AgNO_3 solution is 4.02 \times 10^4 M </em>

<em></em>

<u>Explanation:</u>

The Balanced chemical equation is

1AgNO_3 (aq) +1KCl (aq) > 1 AgCl (s)+1KNO_3 (aq)

Mole ratio of AgNO_3 : KCl is 1 : 1

So moles AgNO_3  = moles KCl

Moles KCl = \frac {mass}{molarmass}

= \frac {0.785 mg}{(39.1+35.5 g per mol)}

= \frac {0.000785 g}{74.6 g  per mol}

= 0. 0000105 mol KCl

= 0.0000105 mol AgNO_3

So  Molarity

= \frac {moles of solute}{(volume of solution in L)}

= \frac {0.0000105 mol}{26.2 mL}

=\frac {0.0000105 mol}{0.0262 L}

= 0.000402M or mol/L is the Answer

(Or) 4.02 \times 10^4 M is the Answer

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