a 10.00ml diluted chloride sample required 44.89 ml of 0.01982 m abno3 to reach the fajans end point. how many moles of cl ion were present in the sample
If you intend to go anywhere in chemistry you need to remember there is a difference between m and M. Did you intend M?
I have no idea what abno3 is.
And I don't buy keyboards not having a caps button.
To find the number of moles of chloride ions (Cl-) in the sample, we need to use the concept of stoichiometry and the equation that relates moles to volume and concentration.
The balanced chemical equation for the reaction between chloride ions (Cl-) and silver nitrate (AgNO3) is:
Cl- + AgNO3 -> AgCl + NO3-
From the equation, we can see that 1 mole of chloride ions reacts with 1 mole of silver nitrate to produce 1 mole of silver chloride.
Given:
Volume of silver nitrate solution (V1) = 44.89 mL = 0.04489 L
Molarity of silver nitrate solution (M1) = 0.01982 M
Volume of chloride sample (V2) = 10.00 mL = 0.01000 L
Using the formula of moles (mols) = Volume (L) x Molarity (M), we can calculate the moles of silver nitrate used (mols of AgNO3).
mols of AgNO3 = V1 x M1
= 0.04489 L x 0.01982 M
= 0.000888 Mols
Since the stoichiometric ratio between AgNO3 and Cl- is 1:1, the moles of chloride ions (mols of Cl-) in the sample will also be 0.000888 moles.
Therefore, there were 0.000888 moles of chloride ions (Cl-) present in the 10.00 mL diluted chloride sample.