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Find the molarity and molality of a 15% solution of H_(2)SO_(4) (density of H_(2)SO_(4)=1.020gcm^(-3)). (Atomic mass : H=1,O=16, S=32 a.m.u.). |
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Answer» Solution :`15%` solution of `H_(2)SO_(4)` means 15 g of `H_(2)SO_(4)` are present in 100 g of the solution, i.e., Mass of `H_(2)SO_(4)` DISSOLVED = 15 g , Mass of the solution = 100 g, Density of the solution = 1.02 `g//cm^(3)` (given) Calculation of molality : Mass of solution = 100g , Mass of `H_(2)SO_(4)` (solute) = 15 g Mass of water (SOLVENT ) = `100-15=85g=(85)/(1000)kg=0.085kg` `"Mass of water (solvent)"=100-15=85g=(85)/(1000)kg=0.085kg` `"Molar mass of "H_(2)SO_(4)="98 g mol"^(-1)""therefore""15gH_(2)SO_(4)=(15g)/(98"g mol"^(-1))=0.153"moles"` `"Molality"=("No. of moles of solute")/("Mass of solvent in kg")=(0.153mol)/(0.085kg)=1.8mol kg^(-1)=1.8m` Calculation of molarity = 15 g of `H_(2)SO_(4)=0.153` moles (CALCULATED above) `"Volume of solution"=("Mass of solution")/("Density of solution")=(100)/(1.02)=98.04cm^(3)=(98.04)/(1000)L=0.09804L` `"Molarity "=("No. of moles of the solute")/("Volume of solution in litres")=(0.153mol)/(0.09804L)=1.56mol L^(-1)=1.56M` |
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