Question

the vapour pressure of pure water is 760 mm at 25 degree Celsius the vapour pressure of solution containing 1 (m) solution of glucose is​

Answer 1

Answer:

746.32 mm

Explanation:

According to the lowering of vapor pressure,

( p° - p ) / p° = W₂ / M₂  *  M₁  / W₁   ......................... (a)

Where,

W₂ --- Weight of the solute

M₂  --- Weight of the solvent

M₁  --- Molecular mass of the solute - 18

W₁ --- Molecular mass of the solvent

p    --- Vapor pressure of the solution

p°  --- Vapor pressure of the pure solvent - 760

Molality = Number of moles in solute / Solvent in Kg

            = ( W₂ / M₂ ) / W₁ in g  *  10³

Therefore,

in the above equation (a)

( p° - p ) / p° = m  *  M₁  ..............................................(b)

( 760 - p ) / 760 = 1 * 10⁻³  * 18

                         = 746.32 mm

Answer 2

Answer:

Given : Molality of glucose solution = 1m = 1 mole is dissolved in 1 kg of solvent

= mass of solute  (glucose)=

= mass of solvent  = 1 kg = 1000 g

= molar mass of solvent (water) = 18 g/mole

= molar mass of solute (glucose)= 180 g/mole

Now put all the given values in this formula ,we get the vapor pressure of the solution.

When a liquid evaporates, the gaseous molecules created escape into the air. If the liquid is in a closed container, the gaseous molecules created will not escape but remain above the liquid. These evaporated particles created a pressure above the liquid, this is known as the vapor pressure.

If a solid is dissolved into the liquid, a solution is created. The vapor pressure of the solution is lowered by the addition of the solute. Raoult’s law explains how the vapor pressure of a liquid is altered by the addition of a solute.

Psolution = the vapor pressure of the solution

solvent = the mole fraction of the solvent in the solution

P°solvent = the vapor pressure of the pure solvent at standard conditions"