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andreyandreev [35.5K]
2 years ago
9

A solution contains 4.08 g of chloroform (CHCl3) and 9.29 g of acetone (CH3COCH3). The vapor pressures at 35 ∘C of pure chlorofo

rm and pure acetone are 295 torr and 332 torr, respectively. Assuming ideal behavior, calculate the vapor pressures of each of the components and the total vapor pressure above the solution.

Chemistry
2 answers:
Sergeeva-Olga [200]2 years ago
8 0

Explanation:

Below is an attachment containing the solution.

Marysya12 [62]2 years ago
5 0

Answer:

Vp chloroform  = 51.92 torr

Vp acetone = 273.57torr

total Vp = 325.49torr

Explanation:

Step 1: Data given

Mass of chloroform = 4.08 grams

Mass of acetone = 9.29 grams

The vapor pressure at 35 °C of chloroform = 295 torr

The vapor pressure at 35 °C of acetone = 332 torr

Step 2: Calculate moles

Pressure of solution = Pressure solvent * mole fraction

mole fraction = moles solvent / total moles

Moles chloroform = mass chloroform / molar mass chloroform

Moles chloroform = 4.08 grams / 119.5g/mole = 0.0341moles  

Moles acetone = 9.29 grams  / 58g/mole = 0.160 moles

Step 3: Calculate mol fractions

Mol fraction chloroform = 0.0341 moles / (0.0341+0.160)moles

Mol fraction chloroform = 0.176

Mol fraction acetone = 0.160 /(0.0341+0.160)

Mol fraction acetone = 0.824

Step 4: Calculate vapor pressure

Vp = pressure  * mole fraction

Vp chloroform  = 295torr * 0.176 = 51.92 torr

Vp acetone = 332 torr * 0.824 = 273.57torr

total Vp = 325.49torr

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C = 17.39 / 12 = 1.45

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C = 1.45 / 1.45 = 1

F = 4.35 / 1.45 = 3

 

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CF3 

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2 years ago
Consider 2.4 moles of a gas contained in a 4.0 L bulb at a constant temperature of 32°C. This bulb is connected to an evacuated
Luden [163]

Complete Question

The complete question is shown on the first uploaded image

Answer

a

As the valve is opened , the gas will flow into the empty container until the both containers have the same pressure

b

\Delta H = 0\ , \Delta E = 0 , q= 0 , w= 0

c

The driving force for this process is the increase in entropy this is because the movement of the internal energy of the gas into a larger volume, what this does is that it increases the amount of disorder(entropy).

Explanation

In order to obtain the parameter in the part B of the question we are first obtain the initial pressure, using the ideal gas equation  

                      P = \frac{nRT}{V}

                     P = \frac{(2.4mol)(0.0821\frac{1 atm}{K \cdot \ mol} )}{4.0L}

                         P =15 \ atm

The next thing is to obtain the new pressure of the gas , using boyle's law

              P_1V_1 = P_2V_2

                  P_2 = \frac{P_1 V_1}{V_2}

                  P_2 = \frac{(15 \ atm)(4.0L)}{24.0 L}

                  P_2 = 2.5 \ atm    

Since the this process is isothermal , the change in heat is equal to zero

                      i.e  q = 0 J

  The workdone to move  the gas to the other container is zero because the  the pressure at this second container is zero due to the fact that it is a vacuum

    i.e  w = -P_{external} \Delta V

              =-(0 \ atm) (24.0 - 4.0L)

              = 0L \cdot atm

  Since the change in heat is zero and the workdone is zero then the change in internal energy is equal to 0

     This is because the change in internal energy is equal to a summation of change in heat and the workdone

                i.e \Delta E = q + w

                            = 0J

Generally the change in enthalpy is mathematically represented as

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Since the temperature is zero this means that the change in temperature is zero , substituting this value for change in temperature into the equation for  change in enthalpy

           \Delta H = n C_p (0)

                   = 0J

 

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cluponka [151]

Answer:

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Explanation:

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ΔT = Kf×m×i

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Molality of 9.04g I₂ (Molar mass: 253.8g/mol) in 75.5g of benzene (0.0755kg) is:

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As freezing point of benzene is 5.5°C, the new freezing point of the solution is:

5.5°C - 2.4°C =

<h3>3.1°C</h3>

<em />

3 0
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