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,

$\frac{p^o-p_s}{p^o}=\frac{w_2M_1}{w_1M_2}$

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.

$\frac{23.76-p_s}{23.76}=\frac{12.25\times 18.02}{176.3\times 342.3}$

$p_s=23.67mmHg$

Therefore, the vapor pressure of solution is 23.67 mmHg.