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GaryK [48]
3 years ago
12

A solution contains 6.21 g of ethylene glycol dissolved in 25.0 g of water. If water has a boiling point elevation constant of 0

.512°C•kg/mol, what is the boiling point of the solution? (molar mass of ethylene glycol = 62.1 g/mol; boiling point of pure water = 100.00°C)
Chemistry
2 answers:
adell [148]3 years ago
7 0
The answer is D.) i just took the quiz
timurjin [86]3 years ago
5 0
Calculate first the number of moles of ethylene glycol by dividing the mass by the molar mass.
                           n = (6.21 g ethylene glycol) / 62.1 g/mol
                              n = 0.1 mol
Then, calculate the molality by dividing the number of moles by the mass of water (in kg).
                           m = 0.1 mol/ (0.025 kg) = 4m
Then, use the equation,
                       Tb,f = Tb,i + (kb)(m)
Substituting the known values,
                       Tb,f = 100°C + (0.512°C.kg/mol)(4 mol/kg)
                          <em>Tb,f = 102.048°C</em>
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How many moles of chlorine gas and 120°C and 33.3 ATM would occupy a vessel of 12 L?
Mashutka [201]
<h3>Answer:</h3>

12.387 moles

<h3>Explanation:</h3>

We are given;

Temperature of chlorine, T = 120°C

But, K = °C + 273.15

Therefore, T = 393.15 K

Pressure, P = 33.3 Atm

Volume, V = 12 L

We are required to calculate the number of moles of chlorine gas,

To find the number of moles we are going to use the ideal gas equation;

PV = nRT

R is the ideal gas constant, 0.082057 L.atm/mol.K

Therefore, rearranging the formula;

n = PV÷RT

Hence;

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  = 12.387 moles

Therefore, the number of moles of chlorine are 12.387 moles

8 0
3 years ago
The constant volume heat capacity of a gas can be measured by observing the decrease in temperature when it expands adiabaticall
miv72 [106K]

Answer:

The value is  C_p  = 42. 8 J/K\cdot mol

Explanation:

From the question we are told that  

     \gamma = \frac{C_p }{C_v}

The  initial volume of the  fluorocarbon gas is  V_1 = V

 The final  volume of the fluorocarbon gas isV_2 = 2V

  The initial  temperature of the fluorocarbon gas is  T_1  =  298.15 K

  The final  temperature of the fluorocarbon gas is T_2  =  248.44 K

   The initial  pressure is P_1  = 202.94\  kPa

    The final   pressure is  P_2  =  81.840\  kPa

Generally the equation for  adiabatically reversible expansion is mathematically represented as

       T_2 =  T_1  * [ \frac{V_1}{V_2} ]^{\frac{R}{C_v} }

Here R is the ideal gas constant with the value  

        R =  8.314\  J/K \cdot mol

So  

   248.44 =   298.15  * [ \frac{V}{2V} ]^{\frac{8.314}{C_v} }

=> C_v  =  31.54 J/K\cdot mol

Generally adiabatic reversible expansion can also be mathematically expressed as

    P_2 V_2^{\gamma} = P_1 V_1^{\gamma}

=>[ 81.840 *10^3] [2V]^{\gamma} = [202.94 *10^3] V^{\gamma}    

=>  2^{\gamma} =  2.56

=>    \gamma =  1.356

So

     \gamma  =  \frac{C_p}{C_v} \equiv  1.356 = \frac{C_p}{31.54}

=>    C_p  = 42. 8 J/K\cdot mol

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