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Likurg_2 [28]
3 years ago
11

A nonvolatile solute is dissolved in benzene so that it has a mole fraction of 0.139. What is the vapor pressure of the solution

? (P° of benzene is 26.5 torr)
Chemistry
1 answer:
lapo4ka [179]3 years ago
7 0

Answer:

The vapor pressure of the solution is 3.69 torr

Explanation:

Step 1: Data given

Mole fraction of benzene in the solution = 0.139

P° of benzene is 26.5 torr

Step 2: Calculate the vapor pressure of the solution

Psolution = Xbenzene * P°benzene

⇒with Psolution = the vapor pressure of the solution

⇒with Xbenzene = the mole fraction of benzene = 0.139

⇒with P°benzene = the vapor pressure of pure benzene = 26.5 torr

Psolution = 0.139 * 26.5 torr

Psolution = 3.69 torr

The vapor pressure of the solution is 3.69 torr

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castortr0y [4]
Molarity = moles / liter

a) M = 2/4 = 0.5 M

b) Moles = 4/(30 + 16 + 1)
= 0.085
M = 0.085 / 2 = 0.0425 M

c) Moles = 5.85 / (23 + 35.5)
= 0.1
M = 0.1 / 0.4
= 0.25 M
3 0
3 years ago
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8.5ml of a sample of sea water solution was added to a 44.317g evaporating dish the combination weighted 52.987g after evaporati
Ratling [72]

Answer:  There is no question, but we can calculate a couple of items:

Density of sea water sample = (52.987g-44.317g)/8.5ml

Inorganic content of sample (mostly salts) = (44.599g-44.317g)/(52.987g-44.317g) x 100% = percent inorganics in water sample

Explanation:

5 0
3 years ago
A 5.00-g sample of copper metal at 25.0 °C is heated by the addition of 133 J of energy. The final temperature of the copper is
vekshin1

<u>Answer:</u> The final temperature of the copper is 95°C.

<u>Explanation:</u>

To calculate the final temperature for the given amount of heat absorbed, we use the equation:

Q= m\times c\times \Delta T

Q = heat absorbed  = +133 J (heat is added to the system)

m = mass of copper = 5.00 g

c = specific heat capacity of copper = 0.38 J/g ° C      

\Delta T={\text{Change in temperature}}=T_2-T_1

T_1=25^oC

Putting values in above equation, we get:

+133J=5.00g\times 0.38J/g^oC\times (T_2-25)\\\\T_2=95^oC

Hence, the final temperature of the copper is 95°C.

3 0
3 years ago
A solution contains 50.0g of heptane (C7H16)and 50.0g of octane (C8H18) at 25 degrees C.The vapor pressures of pure heptane and
AleksandrR [38]

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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