Question

if 65.1 ml of silver nitrate solution reacts with excess potassium chloride solution to yield 0.365 g of precipitate,what is the molarity of silver ion in the original solution

Answer 1

first write the balanced equation

AgNO₃ + KCl -------> AgCl + KNO₃

so precipate formed here is AgCl

no ofmoles of AgCl = weight of AgCl / molar mass of AgCl

= 0.365 g / 143.32 g/mol

= 0.00255 moles

now agin comming to the balanced equation

one mole of AgCl can form from one mole og AgNO3 and one mole of KCl right?

so no of moles of AgCl formed = no of moles of AgNO3 = 0.00255 moles

volume of AgNO3 = 65.1 mL convert in to liters = 0.0651 L

we have volume of AgNO3 and moles of AgNO3

molarity of AgNO3 = moles / volume

= 0.00255 moles / 0.0651 L

= 0.0391 M