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प्रश्न
Derive the expression for molar mass of solute in terms of boiling point elevation of solvent.
उत्तर
The boiling point elevation is directly proportional to the molality of the solution. Thus,
∆Τb = Kb m ….(1)
Suppose we prepare a solution by dissolving W2 g of solute in W1 g of solvent. Moles of solute in W1 g of solvent = `"W"_2/"M"_2`
where, M2 is the molar mass of solute.
Mass of solvent = W1 g = `("W"_1 "g")/(1000 "g"//"kg") = "W"_1/1000`kg
The molality is expressed as,
m = `"Moles of solute"/"Mass of solvent in kg"`
m = `("W"_2//"M"_2 "mol")/("W"_1//1000 "kg")`
m = `(1000 "W"_2)/("M"_2 "W"_1)` mol kg-1 ....(2)
Substituting equation (2) in equation (1), we get,
`Delta "T"_"b" = (1000 "K"_"b" "W"_2)/("M"_2 "W"_1)`
Hence,
`"M"_2 = (1000 "K"_"b" "W"_2)/(Delta "T"_"b" "W"_1)`
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