# If 75.0 mL of silver nitrate solution reacts with excess potassium chloride solution to yield 0.640 g of precipitate, what is the molarity of silver ion in the original solution?

The molarity of silver ion in the original solution can be determined by performing a stoichiometric calculation. The reaction between silver nitrate and potassium chloride is as follows: AgNO3 (aq) + KCl (aq) –> AgCl (s) + KNO3 (aq). In order to determine the molarity of silver ion, we will need to calculate the number of moles of silver nitrate that are present in 75.0mL of solution. This can be done using the following equation:
M = n/V
Where M is molarity, n is moles and V is volume in liters. To convert mL to L, we must divide by 1000; thus, our equation becomes:
M = n/V x 1000
Since 1 mole of anything contains 6.02 x 10^23 molecules, we can use this value and the mass given to find out how many moles were used in the reaction. We know that 0.640 g precipitate was formed from the reaction which means that 0.640 g Silver Chloride was produced from 1 mole Silver Nitrate reacting with 1 mole Potassium Chloride according to our reactants and products above. Thus, since 1 mole has a mass equal to its atomic weight on the periodic table which for Silver Chloride would be 143 g/mol, we now have enough information to solve for our unknowns and complete our stoichiometric calculation:
n=(0.640gAgCl)(1molAgCl/143gAgCl)=4.431×10^-3 mol AgNO3
Once we have calculated how many moles were used in this reaction it’s time to finish up with calculating for Molarity using our equation :
M=(4.431×10^-3molAgNO3)/(75ml/1000L)=5.908Mol*L^-1 or 59080 Mol*L^-1

## If 75.0 mL of silver nitrate solution reacts with excess potassium chloride solution to yield 0.640 g of precipitate, what is the molarity of silver ion in the original solution?

Therefore, based on this calculation, we can conclude that there was an initial concentration of 59080 Mol*L^-1 or 5.908 M Ag+ ions present in 75 mL solution before any chemical reactions occurred between it and potassium chloride resulting ultimately yielding a precipitate consisting primarily of silver chloride particles suspended within water molecules along with some excess solutes consisting mostly likely of excess potassium chloride still dissolved within solution even after precipitation occurred upon mixing both solutions together but this aspect however cannot be accurately measured seeing as no additional measurements were provided other than simply just the amount originally given at firstwhichwas just simply only being 75mL .

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