Question

The vapor pressure of water is 23.76 mm Hg at 25 °C. A nonvolatile, nonelectrolyte that dissolves in water is sucrose. Calculate the vapor pressure of the solution at 25 °C when 12.25 grams of sucrose, C12H22O11 (342.3 g/mol), are dissolved in 176.3 grams of water. water = H2O = 18.02 g/mol.

Answer 1

Answer : The vapor pressure of solution is 23.67 mmHg.

Solution:

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

`(p^o-p_s)/(p^o)=(w_2M_1)/(w_1M_2)`

where,

`p^o` = vapor pressure of pure solvent (water) = 23.76 mmHg

`p_s` = vapor pressure of solution= ?

`w_2` = mass of solute (sucrose) = 12.25 g

`w_1` = mass of solvent (water) = 176.3 g

`M_1` = molar mass of solvent (water) = 18.02 g/mole

`M_2` = molar mass of solute (sucrose) = 342.3 g/mole

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

`(23.76-p_s)/(23.76)=(12.25* 18.02)/(176.3* 342.3)`

`p_s=23.67mmHg`

Therefore, the vapor pressure of solution is 23.67 mmHg.