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

Step-by-step 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,

`(p^o-p_s)/(p^o)=i* x_2`

where,

`(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* volume = 1g/ml* 50000ml=50000g`

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

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

`x_2` = mole fraction of solute

=
`(9)/(2787)=3.2* 10^(-3)`

`(29.96-p_s)/(29.96)=1* 3.2* 10^(-3)`

`p_s=29.86mmHg`

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