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Find the molarity and molality of a 15% solution of H_(2)SO_(4) (density of H_(2)SO_(4)="1.020 g cm"^(-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 = 100 g , Mass of `H_(2)SO_(4)` (solute) = 15 g Mass of water (solvent) `=100-15=85 g =(85)/(1000)kg=0.085kg` Molar mass of `H_(2)SO_(4)="98 g mol"^(-1)""therefore"15 g "H_(2)SO_(4)=(15g)/("98 g mol"^(-1))="0.153 moles"` `"Molality"=("No. of moles of solute")/("Mass of solvent in kg")=("0.153 mol")/("0.085 kg")="1.8 mol 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)="98.04 cm"^(3)=(98.04)/(1000)L=0.09804L` `"Molarity"=("No. of moles of the solute")/("Volume of solution in LITRES")=("0.153 mol")/("0.09804 L")="1.56 mol L"^(-1)="1.56 M"` |
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