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Mama L [17]
3 years ago
11

Given one mole of each substance, which of the following will produce the FEWEST particles in aqueous solution? 1. sodium nitrat

e 2. CH2Cl2 3. K2SO4 4. sodium phosphate
Chemistry
1 answer:
svlad2 [7]3 years ago
3 0

<u>Answer:</u> The substance that produces fewest particles is CH_2Cl_2

<u>Explanation:</u>

Ionization reaction is defined as the reaction in which an ionic compound dissociates into its ions when dissolved in aqueous solution.

Covalent compounds do not dissociate into ions when dissolved in aqueous solution.

For the given options:

  • <u>Option 1:</u>  Sodium nitrate

The chemical formula of sodium nitrate is NaNO_3

The ionization reaction for the given compound follows:

NaNO_3(aq.)\rightarrow Na^+(aq.)+NO_3^-(aq.)

This produces in total of 2 ions.

  • <u>Option 2:</u>  CH_2Cl_2

The given compound is a covalent compound and do not dissociate into its ions. It remains as such as a single unit.

  • <u>Option 3:</u>  K_2SO_4

The chemical name for the given compound is potassium sulfate.

The ionization reaction for the given compound follows:

K_2SO_4(aq.)\rightarrow 2K^+(aq.)+SO_4^{2-}(aq.)

This produces in total of 3 ions.

  • <u>Option 4:</u>  Sodium phosphate

The chemical formula of sodium phosphate is Na_3PO_4

The ionization reaction for the given compound follows:

Na_3PO_4(aq.)\rightarrow 3Na^+(aq.)+PO_4^{3-}(aq.)

This produces in total of 4 ions.

Hence, the substance that produces fewest particles is CH_2Cl_2

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Answer:

a)Pheptane = 24.3 torr          

Poctane = 5.12 torr    

b)Ptotal vapor = 29.42 torr

c)  81 % heptane

    19 % octane

d) See explanation below

Explanation:

The partial pressure is given by Raoult´s law as:

Pa = Xa Pºa where Pa = partial pressure of component A

                               Xa = mole fraction of A

                               Pºa = vapor pressure of pure A

For a binary solution what we have to do is compute the partial  vapor pressure of each component and then add them together to get total vapor pressure.

In order to calculate the composition of the vapor  in part b), we will first calculate the mole fraction of each component in the vapor which is given by the relationship:

          Xa = Pa/Pt where Xa = mol fraction of  in the vapor

                                       Pa = partial pressure of A as calculated above

                                        Pt = total vapor pressure

Once we have mole fractions we can calculate the masses of the components for part c)    

a)                  

 MW heptane = 100.21 g/mol

 MW octane = 114.23 g/mol

mol heptane = 50.0 g / 100.21 g/mol = 0.50 mol

mol octane = 50.0 g/ 114.23 g/mol = 0.44 mol

mol total = 0.94 ⇒ Xa= 0.50/0.94 = 0.53 and

                             Xb= 0.44/0.94 = 0.47

Pheptane = 0.53 x 45.8 torr = 24.3 torr

Poctane = 0.47 x 10.9 torr = 5.12 torr

b) Ptotal = 24.3 torr +5.12 torr = 29.42 torr

c) We will call Y the mole fraction in the vapor to differentiate it from the mole fraction in solution

Y heptane (in the vapor) = 24.3 torr/ 29.42 torr = 0.83

Y octane (in the vapor) = 5.12 torr/ 29.42 torr = 0.17

d) To solve this part   we will assume that since the molecular weights are similar then having a mole fraction for heptane of 0.82, we could say that for every mole of mixture we have 0.82 mol heptane and 0.17 mol octane  and then we can calculate the masses:

0.82 mol x 100.21  g/mol = 82.2 g

0.17 mol x 114.23 g/mol =  19.4 g

total mass = 101.6

% heptane = 82.2 g/101.6g x 100 = 81 %

% octane = 19 %

There is another way to do this more exactly by calculating the average molecular weight of the mixture:

average MW = 0.83 (100.21 g/mol)  + 0.17 ( 114.23 g/mol ) = 102. 6 g/mol

and then  having a mol fraction of 0.83  means in 1 mol of mixture we have 0.83 mol heptane and 0.17 mol octane then the masses are:

mass heptane = 0.83 x 100.21 g/mol = 83.2 g

mass octane = 0.17 x  114.23 g/mol = 19.4 g

mass of mixture = 1 mol x MW mixture = 1 mol x 102.6 g/mol 102.6 g

% heptane = (83.2 g/ 102.6 g ) x 100 g = 81 %

% octane = 100 - 81 = 19 %

d)The composition of the vapor is different from the composition of the solution because the vapor is going to be richer in the more volatile compound in the solution which in this case is heptane ( 45.8  vs 10.9 torr).

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