Question

Molarity: A sample of commercial sulphuric acid is `98% H_(2) SO_(4)` by mass and its specfic gravity is `1.84`. Caculate the molartiy of this sulburic acid solution.
Strategy: The density of a solution (grams per milliliter) is numercially equal to its specific gravity. Thus, the density of the solution is `98% H_(2) SO_(4)` by mass. Thus, every `100g` of soultuon contains `98g` of pure `H_(2) SO_(4)`. From the mass of `H_(2) SO_(4)`. we calculate its moles and from the density , we calculate its volume. Finally, we calculate molartity using its definition.

Answer 1

Step1: Calculate the moles of solure.
Consider `100g` solutiohn, it contains `98g` of pure `H_(2) SO_(4)`. Thus,
Number of moles of `H_(2) SO_(4) = ("Mass of" H_(2) SO_(4))/("Molar mass of" H_(2) SO_(4))`
`= (98.0g)/(98.0g mol^(-1)) = 1.00 mol`
Step 2: Calculate the volume of soltuion.
Volume of `H_(2) SO_(4)` Solution `= ("Mass of" H_(2) SO_(4) "solution")/("Density of" H_(2) SO_(4) "solution")`
`= (100g)/(1.84g mL^(-1))`
`= 54.4 mL = 5.44xx10^(-2) L`
Step 3: Calculate the molarity of the solution
Molarity of `H_(2) SO_(4)` solution `= ("Number of moles of solute")/("Liters of solution")`
`= (1.00 "mol")/(5.44xx10^(-21) L)`
`= 18.4 mol L^(-1) = 18.4M`
Alternatively, in one setup, considering `1 L` of solution,
`(? mol H_(2) SO_(4))/("L soln.") = (1.84 "g soln")/("mL soln.") xx (1000 "mL soln")/("L soln.") xx (98g H_(2) SO_(4))/(100g "soln.")`
`xx (1 "mol" H_(2) SO_(4))/(98g H_(2) SO_(4))`
`= 18.4 M`
Note that a series fo three unit factors is used.