Question

9 moles of Ammonia (NH3) are added to 50 L of H2O at a temperature of 29°C. The vapor pressure of water alone is 29.96 mmHg at 29°C. What is the vapor pressure of the Ammonia solution?

Answer 1

Answer:  The vapor pressure of the solution at $29^0C$ is 29.86 mm Hg

Explanation:

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,

$\frac{p^o-p_s}{p^o}=i\times x_2$

where,

$\frac{p^o-p_s}{p^o}$= relative lowering in vapor pressure

i = Van'T Hoff factor = 1 (for non electrolytes)

$x_2$ = mole fraction of solute  

=$\frac{\text {moles of solute}}{\text {total moles}}$

Given : 9 moles of $NH_3$ are dissolved in 50 L or 50000 ml of water

mass of water = $density\times volume = 1g/ml\times 50000ml=50000g$  

moles of solvent (water) = $\frac{\text{Given mass}}{\text {Molar mass}}=\frac{50000g}{18g/mol}=2778moles$

Total moles = moles of solute  + moles of solvent = 9 mol + 2778 mol = 2787  

$x_2$ = mole fraction of solute  

=$\frac{9}{2787}=3.2\times 10^{-3}$

$\frac{29.96-p_s}{29.96}=1\times 3.2\times 10^{-3}$

$p_s=29.86mmHg$

Thus the vapor pressure of the solution at $29^0C$ is 29.86 mm Hg