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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. |
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Answer» 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. |
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