Question

Raoult's law accounts for the fact that the vapor pressure of a solvent will decrease as the mole fraction of the solvent is decreased. In considering the mole fraction, it is important to consider the total moles of dissolved particles. Remember: a particle can be a dissolved molecule or ion. Which aqueous solutions would have the lowest vapor pressure

Answer 1

Raoult's law accounts for the fact that the vapor pressure of a solvent will decrease as the mole fraction of the solvent is decreased. In considering the mole fraction, it is important to consider the total moles of dissolved particles. Remember: a particle can be a dissolved molecule or ion. Which aqueous solutions would have the lowest vapor pressure.

0.1 M
`NH_4NO_3(aq)` , 0.1 M
`NaF(aq)`, 0.1 M
`LiNO_3(aq)`, 0.1 M
`Na_3PO_4(aq)` and 0.1 M
`HC_2H_3O_2(aq)`

Answer: 0.1 M
`Na_3PO_4`

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

`x_2` = mole fraction of solute

1. For 0.1 M
`NH_4NO_3`

`NH_4NO_3\rightarrow NH_4^(+)+NO_3^(-)`

, i= 2 as it is a electrolyte and dissociate to give 2 ions. and concentration of ions will be
`1* 0.1+1* 0.1=0.2`

2. For 0.1 M
`NaF`

`NaF\rightarrow Na^(+)+F^(-)`

, i= 2 as it is a electrolyte and dissociate to give 2 ions, concentration of ions will be
`1* 0.1+1* 0.1=0.2`

3. For 0.1 M
`LiNO_3`

`LiNO_3\rightarrow Li^(+)+NO_3^(-)`

, i= 2 as it is a electrolyte and dissociate to give 2 ions. and concentration of ions will be
`1* 0.1+1* 0.1=0.2`

4. For 0.1 M
`Na_3PO_4`

`Na_3PO_4\rightarrow 3Na^(+)+PO_4^(3-)`

, i= 4 as it is a electrolyte and dissociate to give 4 ions. and concentration of ions will be
`3* 0.1+1* 0.1=0.4`

5. For 0.1 M
`HC_2H_3O_2(aq)`

`HC_2H_3O_2(aq)\rightarrow CH_3COO^(-)+H^(+)` [/tex]

, i= 2 as it is a electrolyte and dissociate to give two ions, concentration of ions will be
`2* 0.1=0.2`

Thus as concentration of solute is highest for 0.1 M
`Na_3PO_4` , the vapor pressure will be lowest.