How many grams of dissolved potassium iodide are in 550 mL of a 1.10 mol/L solution?

Question

How do you do this with unit analysis?

Answer 1

Atomic weights: K=39 I=127 KI=166

Let the solution be called S

550mLS x 1,10molKI/1000mLS x 166gKI/1molKI = 100g KI

Answer 2

interest or Molarity = Moles/quantity (in L) Moles = Mass (in g)/ Molar mass (in g/mol) --> can in looking molar mass employing a periodic table Moles of AgNO3= (36.0g)/(169.87g/mol) = 0.2120 moles volume = 325 mL = 0.325 L interest = (0.2120 mol)/(0.325 L) = 0.652 mol/L = 0.652 M desire this facilitates!