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

It's 4.02 x
`10^(-4)`

Step-by-step 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

Answer 2

The molarity of the
`AgNO_3` solution is
`4.02 * 10^4 M`

Step-by-step 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 * 10^4 M` is the Answer