Answer:
0.515 g
Explanation:
Acetone (C₃H₆O) has a boiling point of 56.5 °C. How many grams of the acetone vapor would occupy the 250 mL Erlenmeyer flask at 57 °C and 730 mmHg?
Step 1: Given data
Temperature (T): 57°C
Pressure (P): 730 mmHg
Volume (V): 250 mL
Step 2: Convert "T" to Kelvin
We will use the following expression.
K = °C + 273.15 = 57°C + 273.15 = 330 K
Step 3: Convert "P" to atm
We will use the conversion factor 1 atm = 760 mmHg.
730 mmHg × (1 atm/760 mmHg) = 0.961 atm
Step 4: Convert "V" to L
We will use the conversion factor 1 L = 1,000 mL.
250 mL × (1 L/1,000 mL) = 0.250 L
Step 5: Calculate the moles (n) of acetone
We will use the ideal gas equation.
P × V = n × R × T
n = P × V/R × T
n = 0.961 atm × 0.250 L/(0.0821 atm.L/mol.K) × 330 K
n = 8.87 × 10⁻³ mol
Step 6: Calculate the mass corresponding to 8.87 × 10⁻³ moles of acetone
The molar mass of acetone is 58.08 g/mol.
8.87 × 10⁻³ mol × 58.08 g/mol = 0.515 g
The mass of acetone vapor in grams is equal to 0.515 grams.
Given the following data:
Scientific data:
To calculate the mass of acetone vapor in grams:
First of all, we would determine the number of moles by using the ideal gas equation:
[tex]n = \frac{PV}{RT} \\\\n = \frac{0.961 \times 0.250}{0.0821 \times 330}\\\\n = \frac{0.24025}{27.093}[/tex]
n = 0.008868 moles.
For the mass:
[tex]Mass = number \;of \;moles \times molar\;mass\\\\Mass = 0.008868 \times 58.08[/tex]
Mass = 0.515 grams.
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