Question

26. The vapor pressure of water at 20°C is 17.5 torr. The change in the vapor pressure of an
aqueous solution prepared by adding 36.0 grams of glucose (C6HI206) to 14.4 grams of water
IS
(a) 15.0 torr
(b) 17.5 torr
(c) 16.4 torr
(d) 14.0 torr​

Answer 1

Answer : The correct option is, (d) 14.0 torr

Explanation :

Molar mass of water = 18 g/mole

Molar mass of glucose = 180 g/mole

First we have to calculate the moles of water and glucose.

$\text{Moles of water}=\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{14.4g}{18g/mole}=0.8mole$

$\text{Moles of glucose}=\frac{\text{Mass of glucose}}{\text{Molar mass of glucose}}=\frac{36g}{180g/mole}=0.2mole$

Now we have to calculate the vapor pressure of the solution.

Using Rouault's law equation :

$p_s=X_a\times p^o_a$

$p_s=\frac{n_a}{n_a+n_b}\times p^o_a$

where,

$p_s$ = vapor pressure of the solution = ?

$p^o_a$ = vapor pressure of pure solvent (water) = 17.5 torr

$X_a$ = mole fraction of solvent (water)

$n_a$ = moles of solvent (water) = 0.8 mole

$n_b$ = moles of solute (glucose) = 0.2 mole

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

$p_s=\frac{0.8}{0.8+0.2}\times 17.5torr$

$p_s=14.0torr$

Therefore, the vapor pressure of the solution is, 14.0 torr