At STP, 2.24 L of carbon dioxide gas is collected. What is the number of molecules of gas present in this volume?
1 answer:
Considering the definition of STP conditions and Avogadro's number, the number of molecules of gas present in 2.24 L is 6.023×10²² molecules.
<h3>Definition of STP condition</h3>The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.
<h3>Definition of Avogadro's number</h3>Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023×10²³ particles per mole. Avogadro's number applies to any substance.
<h3>Amount of molecules of carbon dioxide gas</h3>At STP, 2.24 L of carbon dioxide gas is collected. So, you can apply the following rule of three: if by definition of STP conditions 22.4 L are occupied by 1 mole of carbon dioxide gas, 2.24 L are occupied by how many moles of carbon dioxide gas?
amount of moles of carbon dioxide gas= 0.1 moles
Finally, 0.1 moles of carbon dioxide gas are collected.
Then you can apply the following rule of three, considering the Avogadro's number: If 1 mole of carbon dioxide gas contains 6.023×10²³ molecules, 0.1 mole of carbon dioxide gas contains how many molecules?
amount of molecules of carbon dioxide gas= (6.023×10²³ molecules × 0.1 moles)÷1 mole
amount of molecules of carbon dioxide gas=6.023×10²² molecules
Finally, the number of molecules of gas present in 2.24 L is 6.023×10²² molecules.
Learn more about
STP conditions:
Avogadro's Number:
#SPJ1
You might be interested in
Answer : The average atomic mass of chlorine is, 32.37 amu
Explanation :
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
As we are given that,
Mass of isotope 1 = 34.97 amu
Percentage abundance of isotope 1 = 75.77 %
Fractional abundance of isotope 1 = 0.7577
Mass of isotope 2 = 36.97 amu
Percentage abundance of isotope 2 = 24.23 %
Fractional abundance of isotope 2 = 0.2423
Now put all the given values in above formula, we get:
Therefore, the average atomic mass of chlorine is, 32.37 amu