Chemistry, asked by haneydavid0407, 1 month ago

When 3.6 gram of non-volatile solute ‘X” is dissolved in 90 g of water at 293 K, its vapour pressure is lowered by 0.4%.Relative lowering in vapour pressure is (Vapour pressure of water at 293 K 17.5mm Hg)​

Answers

Answered by khadijafarooqsultan
1

Answer:

the solut X dissolved is water in it

Answered by AnkitaSahni
0

Given: Mass of solute X dissolved in water = 3.6 g

           Mass of water = 90 g

           Lowering of Vapour Pressure = 0.4%

To Find : Relative lowering of pressure

Solution:

  • The formula for relative lowering of Vapour Pressure (RLVP) is ;

                          RLVP = \frac{P - P_{s} }{P_{s} }                               (1)

Where P = vapour pressure of solvent

           P_{s} = vapour pressure of solution

     

  1. P = 17.5 mm Hg              (given)
  2. P_{s} is 0.4% less than P

P_{s} = P - \frac{4}{1000}P

P_{s} = \frac{996P}{1000}

                Substituting these values in (1)

                   ⇒   RLVP = \frac{P - \frac{996P}{1000} } {\frac{996P}{1000} }

                   ⇒   RLVP = \frac{4P}{996P} = \frac{1}{249}

           Therefore Relative Lowering is 0.004.

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