Question

vapour pressure of water 16.8Hg at certain temperature if the vapour pressure of the solution is 16.78mm. find the molarity of the solution? ​

Answer 1

The molality of the solution is = 0.066

Relative Lowering of vapour pressure is a colligative property.

• Colligative properties depend on the number of solute particles.
• Colligative properties are independent of the nature of the solute.

Lowering of vapour pressure

= p⁰ – p/p⁰ = xₐ

• p⁰ = Vapour pressure of the pure solvent.

• p = Vapour pressure of a solution containing non-volatile solute.bn
• xₐ= mole fraction of solute.

Given:

• p⁰ = 16.8 mm Hg.
• p  = 16.78 mm Hg.
• Molar mass of water = 18g.

          p⁰ – p/p⁰ = (Molar mass of water/1000) x m

              where, m is the molality.

        = (16.80 - 16.78) / 16.8  =  (18/1000) x m

        m = (1000/18) x (16.8 -16.78)/16.8

        m = (1000/18) x 0.02/16.8

        m = 0.066.

molality = 1000 x molarity /(1000xdensity) - molarity x molecular weight

• According to me in the given question instead of molarity, molality is to be calculated.

Answer 2

Answer:

The molality of the solution is 0.66

Explanation:

  GIVEN:

             Vapour pressure of the water is $16.8 Hg$

              The vapour pressure of the solution is $16.78 mm$

TO FIND : MOLALITY OF THE SOLUTION.

      VAPOUR PRESSURE: Vapour pressure is a measure of the tendency of a material to change into the gaseous or vapour state, and it increases with temperature. The temperature at which the vapour pressure at the surface of a liquid becomes equal to the pressure exerted by the surroundings is called the boiling point of the liquid.

      Lowering of vapour pressure: A reduction in the saturated vapour pressure of a pure liquid when a solute is introduced. If the solute is a solid of low vapour pressure, the decrease in vapour pressure of the liquid is proportional to the concentration of particles of solute; i.e. to the number of dissolved molecules or ions per unit volume.

Relative Lowering of vapour pressure is a colligative property.

Colligative properties depend on the number of solute particles.

Colligative properties are independent of the nature of the solute.

Lowering of vapour pressure

= p⁰ – p/p⁰ = xₐ

p⁰ = Vapour pressure of the pure solvent.

p = Vapour pressure of a solution containing non-volatile solute.bn

xₐ= mole fraction of solute.

Given:

p⁰ = 16.8 mm Hg.

p  = 16.78 mm Hg.

Molar mass of water = 18g.

         p⁰ – p/p⁰ = (Molar mass of water/1000) x m

             where, m is the molality.

molality = 1000 x molarity /(1000xdensity) - molarity x molecular weight

               =$(16.80 - 16.78) / 16.8 = (18/1000) *m$

             $m = (1000/18) * (16.8 -16.78)/16.8$

               $m = (1000/18) * 0.02/16.8$

                     $m = 0.066.$

THE MOLALITY OF SOLUTION IS 0.66.

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