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wel
3 years ago
11

Assuming that the density of vinegar is 1.005 g/mL, calculate the molatrity of acetic acid in vinegar from an average of 4.6% fo

r the mass percentage of acetic acid in vinegar.
Chemistry
1 answer:
posledela3 years ago
5 0
Acetic acid is 60.05 grams/mole. In 1 liter of vinegar or 1000 ml  there would be 0.046% of acetic acid = 46 ml x 1.005g/ml = 46.23 grams/60.05 grams= 0.77 moles per litre of vinegar.This then would be the concentration of acetic acid in for example 1 liter of vinegar. 
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Answer:

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b) 30,11 atm

Explanation:

The Ideal Gas Equation is an equation that describes the behavior of the ideal gases:

                                     PV = nRT

where:

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<em>Note: We can express this values with other units, but we must ensure that the units used are the same as those used in the gas constant.</em>

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\frac{Pv}{RT} = 1 + \frac{B}{v}

where:

  • v is the molar volume [L/mol]
  • B is the second virial coefficient [L/mol]
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a) Ideal gas equation:

We convert our data to the adecuate units:

n = 5 moles

V = 3 dm3 = 3 L

T = 25°C = 298°K

We clear pressure of the idea gas equation and replace the data:

PV = nRT ..... P = nRT/V = 5 * 0,08205 * 298/3 =40,75 atm

b) Truncated virial equation:

We convert our data to the adecuate units:

n = 5 moles

V = 3 dm3 = 3 L

T = 25°C = 298°K

B = -156,7*10^-6 m3/mol = -156,7*10^-3 L/mol

We clear pressure of the idea gas equation and replace the data:

\frac{Pv}{RT} = 1 + \frac{B}{v} ...... P = (1 + \frac{B}{v}) \frac{RT}{v}

and v = 3 L/5 moles = 0,6 L/mol

P = (1 + \frac{-156,7*10^{-3} }{0,6} ) \frac{0,08205*298}{0,6} = 30,11 atm

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