Question

The vapour pressure of a dilute aqueous solution of
a non-volatile solute (molar mass 180 g mol-l) at
373 K is 750 mm Hg. Calculate (i) molality (ii) mole
fraction of solute. (Ans.() 0.731 mol kg (ii) 0.013]​

Answer 1

Answer:

25ºC is 639.7 mm Hg and the vapour pressure of a solution of non–volatile solute in benzene at the same ... The boiling point of the solution at 1 atm is. 2 b(H O). K. = 0.52 K kg mol–1

Answer 2

Answer: Molality of solute is 0.731 and its mole fraction is 0.013

Explanation:

Vapour pressure of pure water at 373 K = 1 atm

Vapour pressure of pure water at 373 K = 760 mm Hg

Let the mole fraction of solute be $x$

∴ Mole fraction of water = $1-x$

Vapour pressure of solution = Vapour pressure of solvent + Vapour pressure of solute

The solute is non volatile, hence its vapour pressure is zero.

$750 = 760 (1-x) + 0$

$1-x = \frac{750}{760}$

$x = 1 - \frac{75}{76}$

$x = \frac{1}{76}$

$x = 0.013$

Molality of solute = Moles of solute / Mass of solvent in kg

If we take 1 mole of solution, it will have :

• 0.013 moles of solute
• 0.987 moles of solvent(water)

Molar mass of water = 18 g

Mass of 0.987 moles of water = 0.987 × 18 g = 17.766 g

Mass of 0.987 moles of water = 0.017766 kg

Molality of solute = 0.013/0.017766

Molality of solute = 0.731

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