Question

An aqueous solution of 2% non volatile solute exertes a pressure of 1.004 bar at the normal boiling point of the solvent.What is the molar mass of the solute

Answer 1

. If temp. Increases , k of electrolyte Sol. Increases because

Answer 2

Answer :

• Molecular mass of the solute = 41.35 g/mol.

Step-by-step explanation :

Given that,

• Vapour pressure of solution, $\rm{p_s}$ = 1.004 bar.
• Mass of solute, $\rm{w_2}$ = 2g.
• Mass of solution = 100 g.
• Mass of solvent, $\rm{w_1}$ = 98 g.

Also,

• Vapour pressure of pure water at the boiling point, $\rm{p^{\circ}}$ = 1 atm = 1.013 bar.
• Molecular mass of water, $\rm{M_1}$= 18 g/mol.

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Applying Raoult's law for dilute solution (being 2%),

$\rm{\dfrac{p^{\circ}-p_s}{p^{\circ}}=\dfrac{n_2}{n_1}=\dfrac{w_2/M_2}{w_1/M_1}=\dfrac{w_2}{M_2}\times{}\dfrac{M_1}{w_1}}$

Applying the given values, we get,

$\rm{\dfrac{(1.013-1.004)}{1.013\:bar}=\dfrac{2\:g}{M_2}\times{}\dfrac{18\:g\:mol^{-1}}{98\:g}}$

$\bold{OR}$

$\rm{M_2=\dfrac{2\times{}18}{98}\times{}\dfrac{1.013}{0.009}\:g\:mol^{-1}}$

$\rm{M_2=41.35\:g\:mol^{-1}}$