At 25 o C, an aqueous solution containing 35.0 wt%H 2 SO 4 has a specific gravit

Question

At 25oC, an aqueous solution containing 35.0 wt%H2SO4 has a specific gravity of 1.2563. Aquantity of the 35% solution is needed that contains 195.5kg ofH2SO4. (a) Calculate the required volume (L) of the solution usingthe given specific gravity. (b) Estimate the percentage error that would have resulted ifpure-component specific gravities of H2SO4(SG = 1.8255) and water had been used for the calculation insteadof the given specific gravity of the mixture. At 25oC, an aqueous solution containing 35.0 wt%H2SO4 has a specific gravity of 1.2563. Aquantity of the 35% solution is needed that contains 195.5kg ofH2SO4. (a) Calculate the required volume (L) of the solution usingthe given specific gravity. (b) Estimate the percentage error that would have resulted ifpure-component specific gravities of H2SO4(SG = 1.8255) and water had been used for the calculation insteadof the given specific gravity of the mixture.

Explanation / Answer

35.0 wt% H2SO4 specific gravity = 1.2563 (a) volume of solution which contains 35 gramsH2SO4 = (35/1.2563+75)*10-3L volume of solution which contains 35 gramsH2SO4 = 102.86*10-3 L volume of solution which contains 195.5 KgH2SO4 = 102.86*10-3*195.5*103/35 L volume of solution which contains 195.5 KgH2SO4 =574.5 L (b) pure-component specific gravity ofH2SO4 = 1.8255 volume of solution which contains 35 gramsH2SO4 = (35/1.8255+75)*10-3L volume of solution which contains 35 gramsH2SO4 = 94.1728*10-3 L volume of solution which contains 195.5 KgH2SO4 = 94.1728*10-3*195.5*103/35 L volume of solution which contains 195.5 KgH2SO4 =526 L percentage error = (574.5-526)*100/574.5 percentage error =8.44%

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