Question

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 of the AgNO3 solution?

Answer 1

Answer:

The molarity of the $AgNO_3$ solution is $4.02 \times 10^4 M$

Explanation:

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

Answer 2

Answer:

It's 4.02 x $10^{-4}$

Explanation:

I tried answering what the person above me said, but that was wrong, so I just put a - to the 4 and it worked