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Sorry for the very basic question but its been a very long time since I've taken chemistry and I'm having difficulty solving this problem.

If I titrate something with 1 molar AgNO3, adding 50 ml of AgNO3 solution... How would I calculate how many mols of Ag that equates to?

What I have so far is that 1 M means that there is 0.05 mol of AgNO3 in that 50 ml sample (M = mols/L). So then how many mols of Ag are there in 0.05 mol of AgNO3? Is there just 0.05 mols of Ag in AGNO3 or do I need to somehow take into consideration the proportion of the rest of the elements?

Thank you!!!

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If I titrate something with 1 molar AgNO3, adding 50 ml of AgNO3 solution... How would I calculate how many mols of Ag that equates to?

A 1 molar solution of $\ce{AgNO3}$ is 1 mole per liter. 50 ml is 0.050 liters

$ 1 \dfrac{\text{mole }\ce{AgNO3}}{\text{liter}}\times0.050\text{ liters} = 0.050 \text{ moles }\ce{AgNO3}$

So then how many mols of Ag are there in 0.05 mol of AgNO3? Is there just 0.05 mols of Ag in AGNO3 or do I need to somehow take into consideration the proportion of the rest of the elements?

$0.050 \text{ moles }\ce{AgNO3}$ is correct since there is only one atom of silver per "molecule" of $\ce{AgNO3}$. There would also only be 0.050 moles of N atoms, but 0.150 moles of O atoms in 0.050 moles of $\ce{AgNO3}$.

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Titrate with known concentration of $\ce{KCl}$ and find the how much $\ce{KCl}$ is required:

p ml.X mol $\ce{KCl}$ = q ml Y mol $\ce{Ag}$ is used there for Y mol of Ag = (p ml $\ce{KCl}$ x X molar $\ce{KCl}$ )/ (q ml $\ce{AgNO3}$ used).

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