Question

Which 1 m solution would have the highest vapor pressure at a given temperature? view available hint(s) which 1 solution would have the highest vapor pressure at a given temperature? li2so4 c6h12o6 nac2h3o2 kcl?

Answer 1

First, we have to know that: The vapor pressure depends on the number of moles of the particles as the lower the number of moles of the particles in the solute, the higher the vapor pressure.
So,
1- Li2So4 has  2mol Li + 1Mol So4 = 3 mol solute.
2- C6H12O6 has  1mol solute (It stays as a single molecule)
3- NaC2H3O2 has  1mol Na + 1 mol C2H3O2 = 2 mol solute
4- KCl has  1mol K + 1 mol Cl = 2 mol solute
So your answer is C6H12O6 (glucose) because it has the lowest no.of moles of particles in solute so, it has the highest vapor pressure.

Answer 2

The 1 m solution of $\boxed{{{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}}$ would have the highest vapor pressure.

Further Explanation:

Colligative properties

The properties that depend only on the concentration of solute and not on their identities are termed as colligative properties. The four such properties are listed below:

1. Relative lowering of vapor pressure

2. Elevation in boiling point

3. Depression in freezing point

4. Osmotic pressure

The decrease in the vapor pressure of the solution after the addition of non-volatile solute is called the relative lowering of vapor pressure. It depends on the amount of solute added and therefore it is a colligative property.

The expression for the relative lowering of vapor pressure is as follows:

$\dfrac{{p_1^0 - {p_1}}}{{p_1^0}} = {{\text{x}}_2}$   …… (1)                                                                      

Here,

$p_1^0$ is the pressure of the pure solvent.

$p_1$ is the pressure of the solution.

$\text{x}_2$ is the mole fraction of the solute.

The formula to calculate the mole fraction of solute is as follows:

${{\text{x}}_2} = \dfrac{{{n_2}}}{{{n_1} + {n_2}}}$   …… (2)                                                                  

Here,

$\text{x}_2$ is the mole fraction of solute.

$\text{n}_2$ is the number of moles of solute.

$n_1$ is the number of moles of solvent.

Incorporating equation (2) in equation (1), the modified equation is as follows:

$\dfrac{{p_1^0 - {p_1}}}{{p_1^0}} = \dfrac{{{n_2}}}{{{n_1} + {n_2}}}$   …… (3)                                                            

Equation (3) indicates the direct relationship between the relative lowering of vapor pressure and the moles of solute particles. Higher the number of solute, more will be the lowering of vapor pressure and lower will be the vapor pressure of the solution and vice-versa.

The concentration of each given solution is the same (1 m).

The dissociation reaction of ${\text{L}}{{\text{i}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is as follows:

 ${\text{L}}{{\text{i}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \rightleftharpoons 2{\text{L}}{{\text{i}}^ + } + {\text{SO}}_4^{2 - }$

Here, one mole of ${\text{L}}{{\text{i}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ dissociates to form two moles of ${\text{L}}{{\text{i}}^ + }$ and one mole of ${\text{SO}}_4^{2 - }$ so three moles of solute are produced.

The dissociation reaction of ${\text{Na}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{3}}}{{\text{O}}_{\text{2}}}$ is as follows:

 

Here, one mole of ${\text{Na}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{3}}}{{\text{O}}_{\text{2}}}$ dissociates to form one mole of ${\text{N}}{{\text{a}}^ + }$ and one mole of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - }$ so two moles of solute are produced.

The dissociation reaction of KCl is as follows:

${\text{KCl}} \rightleftharpoons {{\text{K}}^ + } + {\text{C}}{{\text{l}}^ - }$

Here, one mole of KCl dissociates to form one mole of $\text{K}^+$  and one mole of $\text{Cl}^-$ so two moles of solute are produced.

${{\text{C}}_{\text{6}}}{{\text{H}}_{12}}{{\text{O}}_6}$ is a nonelectrolyte so it cannot dissociate into ions and therefore it has only one mole of solute in it.

Since ${{\text{C}}_{\text{6}}}{{\text{H}}_{12}}{{\text{O}}_6}$ has the least moles of solute (1 mol) so its solution has the highest vapor pressure at a given temperature.

Learn more:

1. Choose the solvent that would produce the greatest boiling point elevation: brainly.com/question/8600416
2. What is the molarity of the stock solution? brainly.com/question/2814870

Answer details:

Grade: Senior School

Chapter: Colligative properties

Subject: Chemistry

Keywords: colligative properties, relative lowering of vapor pressure, n1, n2, x2, p1, C6H12O6, KCl, CH3COO-, K+, Na+, Cl-, NaC2H3O2, Li2SO4, Li+, SO42-, highest vapor pressure, solute, non-volatile.