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Find the molarity and molality of a 15% solution of H2S04 (density of H_2SO_4 solution = 1.10 g cm^(-3), Molecular mass of H_2SO_4 = 98). |
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Answer» Solution :15% solution of `H_2SO_4` means 15 g of `H_2SO_4` are present in 100 g of the solution. Thus, mass of `H_2SO_4` dissolved = 15 g mass of the solution = 100 g. Density of the solution (GIVEN) = `1.10 g cm^(-3)` Calculation of molarity: Moles of `H_(2)SO_(4)` dissovled = `15/98 = 0.135` `therefore` Density `=("Mass")/("Volume")` `therefore` Volume = `("Mass")/("Density")` `therefore` The volume of 100 g of the solution `=100/(1.10)= 90.91 cm^(3) = 0.091 L` The molarity of the solution `=("No. of moles of" H_(2)SO_(4))/("Volume in litres")` `=0.153/0.091 = 1.68 MOL L^(-1)` Hence, the molarity of the given solution is 1.68 M. Calculation of molality : Moles of `H_2SO_4` dissolved = 0.153 Mass of water present in 100 g of solution =100-15 = 85 g = 0.085 kg Molality of the solution `=("No. of moles of" H_(2)SO_(4))/("Mass of water in kg")` `=0.153/0.085 = 1.8 mol kg^(-1)` Hence, the molality of the given solution is 1.8 m |
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