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Calculate the molality of 1 litre soluti...

Calculate the molality of 1 litre solution of 93% `H_(2)SO_(4) ("weight"//"volume")`. The density of solution is 1.84 g `mL^(-1)`.

Text Solution

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Calculation of mass of the solvent (water)
Volume of solution =1L=1000mL
` "Density of solution"=1.84g mL^(-1)`
Mass of solution=`Vxxd=(1000mL)xx(1.84gmL^(-1))=1840g`
`"Mass of "H_(2)SO_(4)//L" of solution"=(93g)/((100mL))xx(1000mL)=930g`
`"Mass of solvent (water)"=1840-930=910g`
Calculation of molality of the solution.
`"Molality (m)"=("Mass of" H_(2)SO_(4)//"Molar mass of" H_(2)S_(4))/("Mass of water in kg")`
`"Mass of "H_(2)SO_(4)=930g,Molar mass of H_(2)SO_(4)=98 g mol^(-1)`
`"Mass of water"=(910g)/1000=0.91 kg `
` "Molality "(m)=((930g//98gmol^(-1)))/((0.91kg))=10.43 mol kg ^(-1)=10.43 m`
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