Chemistry: 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?

If 26.2 ml of agno3 is needed to precipitate all

bhenretty2458

04.01.2022

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

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briannwosu8606

04.01.2022

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

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Deavionaaaaa

27.05.2023

40 cm3

Explanation:

brainliest pls

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