Which of the following represents 1 mole of diborane gas, B2H6?
a. 22.4 L diborane gas at STP
b. 27.68 g diborane gas
c. 6.02 × 1023 diborane gas molecules
d. all of the above
e. none of the above
The correct answer and explanation is :
The correct answer is d. all of the above.
Here’s a breakdown of each choice:
- 22.4 L diborane gas at STP:
At Standard Temperature and Pressure (STP), 1 mole of any ideal gas occupies 22.4 L. This is a direct application of the Ideal Gas Law (PV = nRT), where 1 mole of gas at 0°C (273.15 K) and 1 atm pressure has a volume of 22.4 L. Since diborane gas is treated as an ideal gas under these conditions, it follows this rule, so 1 mole of diborane gas will occupy 22.4 L at STP. - 27.68 g diborane gas:
To find the molar mass of diborane (B2H6), we need to sum the atomic masses of boron (B) and hydrogen (H). Boron has an atomic mass of approximately 10.81 g/mol, and hydrogen has an atomic mass of 1.008 g/mol. The molar mass of diborane is:
[
\text{Molar mass of B2H6} = 2(10.81) + 6(1.008) = 21.62 + 6.048 = 27.668 \, \text{g/mol}
]
Thus, 1 mole of diborane weighs approximately 27.68 grams. - 6.02 × 10²³ diborane gas molecules:
This is the definition of 1 mole of any substance according to Avogadro’s number. 1 mole of any substance contains 6.02 × 10²³ particles (atoms, molecules, etc.), so 1 mole of diborane will contain 6.02 × 10²³ molecules of B2H6.
Since all three choices describe 1 mole of diborane in different ways (volume, mass, and number of molecules), the correct answer is d. all of the above.
