For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2ºC. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take Kb = 0.76 K kg mol

Engineering

Chemistry

Electrolytes and Types

Question

For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2ºC. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is

(take K_b = 0.76 K kg mol^–1)

736

724

710

718

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$∆$ T_b = K_b . m

$or, 2 = 0 .76 \times \frac{2 .5 \times 1000}{M \times 100} \Rightarrow M = \frac{0 .76 \times 25}{2}$

$Now, \frac{P ° - P}{P °} = \frac{n_{solute}}{n_{solvent}} (n_{solvent} > > n_{solute})$

$or, \frac{760 - P}{760} = \frac{(\frac{2 .5}{M})}{(\frac{100}{18})} \Rightarrow P = 724 mm Hg$

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