Choose the aqueous solution with the lowest vapor pressure. these are all solutions of nonvolatile solutes and you should assum ideal van't hoff factors where applicable.
a. 120 m c2h602
b. 040 m (nh4)2so4
c. 060 m li2co3
d. 030 m rbc2h3o2
e. they all have the same vapor pressure

Answer : The correct option is, (C) 0.60 M

Explanation :

According to the relative lowering of vapor pressure, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component of the solution multiplied by the vapor pressure of that component in the pure state.

Lowest vapor pressure ∝ i × m

where,

i = Van't Hoff factor

m = molarity

As we know that the vapor pressure depends on the molarity and the Van't Hoff factor.

So, the higher the product of Van't Hoff factor  and molarity the lower the resultant vapor pressure.

(A) The dissociation of 0.120 M is not possible because it is a non-electrolyte solute. So, the Van't Hoff factor will be, 1.

Lowest vapor pressure = 1 × 0.120 = 0.120 m

(B) The dissociation of 0.40 M will be,

So, Van't Hoff factor = Number of solute particles = = 2 + 1 = 3

Lowest vapor pressure = 3 × 0.40 = 1.2 m

(C) The dissociation of 0.60 M will be,

So, Van't Hoff factor = Number of solute particles = = 2 + 1 = 3

Lowest vapor pressure = 3 × 0.60 = 1.8 m

(c) The dissociation of 0.30 M will be,

So, Van't Hoff factor = Number of solute particles = = 1 + 1 = 2

Lowest vapor pressure = 2 × 0.30 = 0.6 m

From this we conclude that, 0.60 M has the highest product of Van't Hoff factor  and molarity which means that the solution will exhibit the lowest vapor pressure.

Hence, the correct option is, (C) 0.60 M

The aqueous solution with the lowest vapor pressure : 0.060 m Li₂CO₃

Solution properties are the properties of a solution that don't depend on the type of solute but only on the concentration of the solute.

Solution properties of electrolyte solutions differ from non-electrolyte solutions because electrolyte solutions contain a greater number of particles because electrolytes break down into ions. So the Solution properties of electrolytes is greater than non-electrolytes.

The term is used in the Solution properties

that is, the number of moles of solute in 1 kg of solvent

the ratio of the number of moles of solute to the mole of solution

Vapor pressure depends on the mole fraction of the components in the solution

P = Xs. P °

P = vapor pressure solution

P ° = pure vapor pressure

Xs = mole fraction solvent

ΔP = P ° - P where

ΔP = change in vapor pressure

It is assumed that the solvent we use is water with an amount of 1 L or 1 kg, then the concentration of the solutions above is the mole of each solution

From the equation for determining the vapor pressure solution, shows a Proportional Comparisons / Directly proportional with the solvent mole fraction (in this case water)

mole solvent fraction:

Because the mole fraction used in the search for vapor pressure is the solvent mole fraction, the greater the mole solute, the smaller the solvent mole fraction, so the smaller the vapor pressure of solution

For electrolyte solutions, i = van't Hoff factor is calculated

n = number of ions produced by the solution

a = degree of electrolyte ionization

So for electrolyte solutions, the solvent mole fraction becomes:

So that the greater multiplication between mole solute x i , the larger the vapor pressure decreases

A. 0.120 m C₂H₆0₂

C₂H₆0₂: non electrolyte solution, i = 1

mol solute x i = 0.12 .1 = 0.12

For choices B, C and D, which are electrolyte solutions, because α is unknown, the value of i depends on the number of ions produced (n)

B. 0.040 m (NH₄)₂SO₄

(NH₄)₂SO₄ ---> 2NH₄ + SO₄²⁻, n = 3

mole solute x 3 = 0.040 x 3 = 0.12

C. 060 m Li₂CO₃

Li₂CO₃ ---> 2Li⁺ + CO₃²⁻, n = 3

mole solute x 3 = 0.06 x 3 = 0.18

D. 030 m RbC₂H₃O₂

RbC₂H₃O₂ ---> Rb⁺ +  C₂H₃O²⁻, n = 2

mole solute x 2 = 0.03 x 2 = 0.06

The largest mole solute x i value is in Li₂CO₃ solution

colligative properties  

link  

Raoult's law  

link  

The vapor pressure of benzene  

link  

The freezing point of a solution  

link  

link  

link  

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

there are no statements provided but here is why:

sulfur belongs to group 16 because it can be found in nature whether it is in a free or combined state.

hope this