Molarity: A solution contains `3.65g` of hydrogen chloride `(HCl)` in `2.00L` of sollution. Calculate the molarity `(M)` of the solution.
Strategy: Convert grams of `HCl` to moles of `HCl` and then apply Eq(1.20).

Explanation

Step 1: Calculate the moles of solute `(HCl):`
Number of moles of `HCl (n_(HCl)) = ("Mass Of HCl")/("Molar mass of HCl")`
`n_(HCl) = (3.65g)/(36.5g mol^(-1)) = 1.100 mol HCI`
Step 2: Apply the definition of molartiy using Eq (1.20)
`M = ("Molar of solute")/("Liners of soln".) = 0.050 "mol" HCl L^(_1) soln`
or `0.50 "mol" L^(-1)`
Alternatelty, in one setup,
`(? "mol" HCl)/(L soln) = (3.65 g HCl)/(2.00 "L soln") xx (1 "mol" HCl)/(36.5 g HCl) = 0.050 "mol" L^(-1)`