2g each of solutes A and B are dissolved separately in 50g each of the same solvent. Which will show greater elevation in boiling point?

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Solution

A) B solute Depression in freezing point = Kf x m (molality = no. of moles dissolved in per Kg solvent)
so, greater the molar mass lesser the no. of moles so lesser is the depression in freezing points so, B with lower molecular mass will have higher depreesion in freezing point
B)B (Solute) as The similar formula is used for elevation (increase in boiling point)
Tb = Kb x m So,lesser the molar mass higher will be no .of moles and so greater is elevation in boiling point.
C) B solute will show lower vapor pressure because we know that,
P0 -Ps/P0
= W2 x M1 / W1 x M2 Here, lowering of vapor pressure (p0-ps) is directly related with amount in gram and inversely related with molar mass of non-volatile solute. Here molar mass of A is greater than B, so B will have lower vapor pressure.

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