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  • 4 years ago

a solution of H2SO4 with a molal concentration of 5.25m has a density of 1.266 g/ml. what is the molar concentration of this solution?

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  1. anonymous
    • 4 years ago
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    Molal concentration is modeled by this equation...\[b_i=\frac{n_i}{m_{solvent}}\]Where... b_i is the molal concentration of the solution n_i is the number of mols of solute, and m_solvent is the mass of the solvent in kilograms If we assume 1kg of solution... \[5.25m=\frac{n_i}{1kg} \rightarrow n_i=5.25mol\ H_2SO_4\]This means we have 5.25 moles of sulfuric acid for every kg of solvent. Now, we can convert this to mass of sulfuric acid per kg of solvent...\[5.25mol\ H_2SO_4*\frac{98.08g\ H_2SO_4}{1mol\ H_2SO_4}=514.91g\ H_2SO_4\]So we have that many grams of sulfuric acid per kg of solvent. Next, let's look at the general form of the density equation...\[D=\frac{m}{V}\]In our case, we are given the density of the entire solution, which is going to be...\[D=\frac{m_{solute}+m_{solvent}}{V_{solute}+V_{solvent}}\]In our case, the volume of the solute is negligible compared to the volume of the solution. Therefore, we can assume that that value is 0. Now, let's go ahead and plug in some values...\[1.266\frac{g}{mL}=\frac{514.91g+1000g}{V_{solution}}\]Now if we solve for V_solution, we get...\[V_{solution}=1196.61mL\]Now, if we assume that the molar concentration of a solution is modeled by this equation...\[c_i=\frac{n_i}{V_{solution}}\]Where the volume of the solution must be in liters. Next, we can substitute in the number of moles we have in this arbitrary amount of solution (which is 1kg of solvent, or 1.19661mL of solution) as well as the volume of the solution, and we should got a molar concentration value. Let's try it out...\[c_i=\frac{5.25mol\ H_2SO_4}{1.19661L}=4.39M\]

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