Derive the expression for molar mass of solute in terms of boiling point elevation of solvent

Q: What is boiling point derive an expression for elevation in boiling point?

It can be calculated as Kb = RTb2M/ΔHv, where R is the gas constant, and Tb is the boiling temperature of the pure solvent [in K], M is the molar mass of the solvent, and ΔHv is the heat of vaporization per mole of the solvent.

The boiling point elevation is directly proportional to the molality of the solution. Thus,

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How is the molar mass of a solute determined by boiling point elevation?
What is boiling point derive an expression for elevation in boiling point?
What is the relation between elevation in boiling point and molality?

∆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)`

How is the molar mass of a solute determined by boiling point elevation?

Since 0.0869 moles of the solute has a mass of 10 grams, 1 mole of the solute will have a mass of 10/0.0869 grams, which is equal to 115.07 grams. Therefore, the molar mass of the solute is 115.07 grams per mole. ... Boiling Point Elevation Formula..

What is boiling point derive an expression for elevation in boiling point?

It can be calculated as K_b = RT_b²M/ΔH_v, where R is the gas constant, and T_b is the boiling temperature of the pure solvent [in K], M is the molar mass of the solvent, and ΔH_v is the heat of vaporization per mole of the solvent.